International Baccalaureate · DP 1: Analysis and Approaches SL

Quadratic Functions

⌚ ~194 min ▢ 22 blocks ✓ 77 exercises ➣ Prerequisites : Straight Lines, Equations

I Quadratic Functions and Their Forms

I.1 Standard Form

Recappreviously studied — click to reread

Definition — Quadratic Function

A quadratic function is a function that can be written in the form $$f(x) = ax^2 + bx + c$$ where $a, b,$ and $c$ are real numbers and $\boldsymbol{a \neq 0}$.

Skills to practice

Standard Form

I.2 Vertex Form

Recappreviously studied — click to reread

Definition — Vertex Form

The vertex form of a quadratic function is:$$ f(x) = a(x-h)^2 + k $$ where $a, h,$ and $k$ are real numbers and $\boldsymbol{a \neq 0}$.

Proposition — Vertex Form of a Quadratic Function

Every quadratic function $f(x) = ax^2 + bx + c$ (with $a \neq 0$) can be written in vertex form $$f(x) = a(x - h)^2 + k \qquad \text{with} \qquad h = -\frac{b}{2a} \quad \text{and} \quad k = f(h) = c - \frac{b^2}{4a}.$$ This writing is unique: $h$ and $k$ are the coordinates of the vertex of the parabola.

Proof

We reuse completing the square (Proposition “Completing the Square”): since $a \neq 0$, we take $a$ out as a common factor of the first two terms and apply $x^2 + 2\lambda x = (x + \lambda)^2 - \lambda^2$ with $\lambda = \dfrac{b}{2a}$: $$\begin{aligned} f(x) &= a\left(x^2 + \frac{b}{a}x\right) + c \\ &= a\left[\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2}\right] + c && \text{(completing the square)} \\ &= a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c. \end{aligned}$$ This is $a(x - h)^2 + k$ with $h = -\dfrac{b}{2a}$ and $k = c - \dfrac{b^2}{4a}$. Replacing $x$ by $h$ gives $f(h) = a \times 0 + k = k$. Uniqueness: if $f(x) = a(x - h')^2 + k'$ for all $x$, then expanding gives $-2ah' = b$ and $ah'^2 + k' = c$, so $h' = h$ and $k' = k$.

Notation

In many textbooks the vertex form (also called completed-square form) is written $f(x) = a(x - \alpha)^2 + \beta$: the letters $\alpha$ and $\beta$ play exactly the roles of $h$ and $k$, the coordinates of the vertex. In simple cases, the vertex form is obtained directly from the identity $x^2 + 2\lambda x = (x + \lambda)^2 - \lambda^2$.

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Example

Write in vertex form:

$f(x) = x^2 + 6x - 1$, using the identity $x^2 + 2\lambda x = (x + \lambda)^2 - \lambda^2$;
$g(x) = -2x^2 + 8x - 3$, using $h = -\dfrac{b}{2a}$ and $k = g(h)$.

Answer

$x^2 + 6x = x^2 + 2 \times 3 \times x = (x + 3)^2 - 9$, so $$f(x) = (x + 3)^2 - 9 - 1 = (x + 3)^2 - 10.$$ Here $h = -3$ and $k = -10$.
$a = -2$, $b = 8$, $c = -3$: $h = -\dfrac{8}{2 \times (-2)} = 2$ and $k = g(2) = -8 + 16 - 3 = 5$, so $$g(x) = -2(x - 2)^2 + 5.$$ Check: $-2(x - 2)^2 + 5 = -2(x^2 - 4x + 4) + 5 = -2x^2 + 8x - 3$.

Method — Converting from Standard to Vertex Form

There are two common methods to convert a quadratic function from standard form, $f(x)=ax^2+bx+c$, to vertex form, $f(x)=a(x-h)^2+k$.

Method 1: Completing the Square
1. Take out the coefficient $a$ as a common factor of the $x^2$ and $x$ terms: $a(x^2 + \frac{b}{a}x) + c$.
2. Inside the brackets, add and subtract the square of half the coefficient of $x$: $a\left(x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2\right) + c$.
3. Factorise the perfect square: $a\left(\left(x + \frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2\right) + c$.
4. Distribute $a$ to the constant term and simplify to get the form $a(x-h)^2 + k$.
Method 2: Using the Vertex Formula
1. Find the x-coordinate of the vertex using the formula: $\boldsymbol{h = -\frac{b}{2a}}$.
2. Find the y-coordinate of the vertex by substituting $h$ back into the function: $\boldsymbol{k = f(h)}$.
3. Substitute the values of $a$, $h$, and $k$ into the vertex form: $f(x) = a(x-h)^2 + k$.

Example

Convert $f(x) = 2x^2 - 12x + 19$ to vertex form.

Answer

Method 1: Completing the Square$$ \begin{aligned}f(x) &= 2(x^2 - 6x) + 19 && \color{gray}\text{(Take out the common factor 2)} \\ &= 2(x^2 - 6x + 9 - 9) + 19 && \color{gray}\text{(Add and subtract } (6/2)^2 = 9) \\ &= 2\left((x-3)^2 - 9\right) + 19 && \color{gray}\text{(Factorise the perfect square)} \\ &= 2(x-3)^2 - 18 + 19 && \color{gray}\text{(Distribute 2)} \\ &= \boldsymbol{2(x-3)^2 + 1}\end{aligned} $$
Method 2: Using the Vertex Formula
For $f(x)=2x^2 - 12x + 19$, we have $a=2, b=-12, c=19$.
1. Find h: $h = -\dfrac{b}{2a} = -\dfrac{-12}{2(2)} = \dfrac{12}{4} = 3$.
2. Find k: $k = f(3) = 2(3)^2 - 12(3) + 19 = 2(9) - 36 + 19 = 18 - 36 + 19 = 1$.
3. Write in vertex form: With $a=2, h=3, k=1$, the function is $\boldsymbol{f(x) = 2(x-3)^2 + 1}$.

Skills to practice

Converting Standard Form to Vertex Form

I.3 Factored Form

Recappreviously studied — click to reread

Definition — Factored Form

The factored form of a quadratic function, when it exists, is:$$ f(x) = a(x-p)(x-q) $$ where $a, p,$ and $q$ are real numbers and $\boldsymbol{a \neq 0}$.

Method — Converting from Standard to Factored Form

To convert from standard form to factored form, find the roots of the function by solving $f(x)=0$. If the roots are $p$ and $q$, the factored form is $f(x)=a(x-p)(x-q)$. The value of $a$ is the same as in the standard form.

Example

Convert $f(x) = 2x^2 - 8x + 6$ to factored form.

Answer

We solve $2x^2 - 8x + 6 = 0$. First, divide by 2: $x^2 - 4x + 3 = 0$.
Two numbers with product $3$ and sum $-4$ are $-1$ and $-3$, so $x^2 - 4x + 3 = (x - 1)(x - 3)$, and by the null factor law the roots are $p = 1$ and $q = 3$.
The factored form is $\boldsymbol{f(x) = 2(x-1)(x-3)}$.

Skills to practice

Converting Standard Form to Factored Form

II Graphs of Quadratic Functions

II.1 The Base Parabola: $y\equal x^2$

Recappreviously studied — click to reread

Proposition — Key Features

The graph of the function $f(x)=x^2$ is a parabola with the following properties:

It opens upwards (it is concave up).
Its axis of symmetry is the line $x = 0$.
Its vertex is at the point $(0, 0)$.

Skills to practice

The Base Parabola: $y\equal x^2$

II.2 Vertex Form and Key Features

Recappreviously studied — click to reread

Proposition — Key Features from Vertex Form

The graph of $y=a(x-h)^2 + k$ is a parabola with:

Direction of opening: Determined by the sign of $a$.
- If $a > 0$, it opens upwards (concave up).
- If $a < 0$, it opens downwards (concave down).
Axis of Symmetry: The line $x = h$.
Vertex: The point $(h, k)$.

Proof

The graph of $y=a(x-h)^2+k$ is obtained by transforming the graph of $y=x^2$ as follows:

Horizontal translation by h: $y=(x-h)^2$. This moves the vertex from $(0,0)$ to $(h,0)$.
Vertical stretch by a: $y=a(x-h)^2$. The direction of opening now depends on the sign of $a$.
Vertical translation by k: $y=a(x-h)^2+k$. This moves the vertex from $(h,0)$ to $(h,k)$.

Thus, the final vertex is $(h,k)$ and the axis of symmetry is $x=h$.

Proposition — Variations of a Quadratic Function

Let $f(x) = ax^2 + bx + c = a(x - h)^2 + k$ with $a \neq 0$.

If $a > 0$, $f$ is decreasing on $(-\infty, h]$ and increasing on $[h, \infty)$: $f$ has a minimum, equal to $k$, reached at $x = h$.
If $a < 0$, $f$ is increasing on $(-\infty, h]$ and decreasing on $[h, \infty)$: $f$ has a maximum, equal to $k$, reached at $x = h$.

Proof

Let $u$ and $v$ be two real numbers with $u < v$. Using the vertex form, $$f(u) - f(v) = a\big[(u - h)^2 - (v - h)^2\big] = a\,(u - v)\,(u + v - 2h)$$ by the identity $A^2 - B^2 = (A - B)(A + B)$ with $A = u - h$ and $B = v - h$. Since $u < v$, the factor $u - v$ is negative.

If $u < v \le h$, then $u + v - 2h < 0$: the product $(u - v)(u + v - 2h)$ is positive, so $f(u) - f(v)$ has the sign of $a$. For $a > 0$, $f(u) > f(v)$: $f$ is decreasing on $(-\infty, h]$. For $a < 0$, $f(u) < f(v)$: $f$ is increasing there.
If $h \le u < v$, then $u + v - 2h > 0$: the product is negative, so $f(u) - f(v)$ has the sign of $-a$. For $a > 0$, $f$ is increasing on $[h, \infty)$; for $a < 0$, $f$ is decreasing there.

Finally $f(x) - k = a(x - h)^2$ has the sign of $a$ and vanishes only at $x = h$: $k = f(h)$ is the minimum of $f$ when $a > 0$, its maximum when $a < 0$.

Example

Let $g(x) = -2x^2 + 8x - 3$. Give the variation table of $g$ and its extremum.

Answer

$a = -2 < 0$, $h = -\dfrac{8}{2 \times (-2)} = 2$ and $k = g(2) = 5$ (vertex form $g(x) = -2(x - 2)^2 + 5$). Since $a < 0$, $g$ is increasing on $(-\infty, 2]$ and decreasing on $[2, \infty)$; its maximum is $5$, reached at $x = 2$.

Skills to practice

Vertex Form and Key Features

II.3 Factored Form and $x$-Intercepts

Recappreviously studied — click to reread

Proposition — Key Features from Factored Form

The graph of a quadratic function $f(x)=a(x-p)(x-q)$ is a parabola with:

Direction of opening: Determined by the sign of $a$.
- If $a > 0$, it opens upwards (concave up).
- If $a < 0$, it opens downwards (concave down).
$x$-intercepts: The graph meets the $x$-axis at $(p, 0)$ and $(q, 0)$.
- If $p \neq q$, there are two distinct $x$-intercepts.
- If $p=q$, there is only one $x$-intercept at $(p,0)$, which is also the vertex. The graph touches the x-axis at this point.
Axis of Symmetry: The line $x = \frac{p+q}{2}$.
Vertex: The point $(\frac{p+q}{2},f(\frac{p+q}{2}))$.

Proof

x-intercepts: The x-intercepts are found by solving $f(x)=0$. $$ a(x-p)(x-q)=0 $$ By the null factor law, this is true if $x-p=0$ or $x-q=0$. Thus, the roots are $\boldsymbol{x=p}$ and $\boldsymbol{x=q}$.
Axis of Symmetry and Vertex: An axis of symmetry at $x=h$ requires that the function has the same value for any horizontal distance $d$ to the left or right of the axis. Algebraically, this means $f(h-d) = f(h+d)$ for any $d$.
We test the proposed axis of symmetry $h=\frac{p+q}{2}$. We need to show that $f\left(\frac{p+q}{2} - d\right) = f\left(\frac{p+q}{2} + d\right)$.
- Left side: $$ \begin{aligned} f\left(\frac{p+q}{2} - d\right) &= a\left(\left(\frac{p+q}{2} - d\right) - p\right)\left(\left(\frac{p+q}{2} - d\right) - q\right) \\ &= a\left(\frac{q-p}{2} - d\right)\left(\frac{p-q}{2} - d\right) \\ &= a\left(-\left(\frac{p-q}{2} + d\right)\right)\left(\frac{p-q}{2} - d\right) \\ &= -a\left(\left(\frac{p-q}{2}\right)^2 - d^2\right) \end{aligned} $$
- Right side: $$ \begin{aligned} f\left(\frac{p+q}{2} + d\right) &= a\left(\left(\frac{p+q}{2} + d\right) - p\right)\left(\left(\frac{p+q}{2} + d\right) - q\right) \\ &= a\left(\frac{q-p}{2} + d\right)\left(\frac{p-q}{2} + d\right) \\ &= a\left(-\left(\frac{p-q}{2} - d\right)\right)\left(\frac{p-q}{2} + d\right) \\ &= -a\left(\left(\frac{p-q}{2}\right)^2 - d^2\right) \end{aligned} $$
Since both sides are equal, the line $\boldsymbol{x = \frac{p+q}{2}}$ is the axis of symmetry. The vertex must lie on this line, so its x-coordinate is $\frac{p+q}{2}$.

Skills to practice

Finding Key Features from Factored Form

II.4 Axis of Symmetry and Vertex

Recappreviously studied — click to reread

The Proposition “Vertex Form of a Quadratic Function” places the vertex at $h = -\frac{b}{2a}$; a horizontal line shows it on the graph. On a set of axes, cut the parabola $y = ax^2 + bx + c$ (with $a \neq 0$) by a horizontal line $y = m$: whenever the line meets the curve twice, the two $x$-coordinates are the two solutions of $ax^2 + bx + c = m$, and their midpoint has the $x$-coordinate $h$, whatever the height $m$. The vertical line $x = h$ is the axis of symmetry of the parabola, and the vertex is the point where this axis meets the curve.

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Proposition — Axis of Symmetry and Vertex

On a set of perpendicular axes with the same unit on both, let $\mathcal{P}$ be the parabola of equation $y = ax^2 + bx + c$ with $a \neq 0$, and let $h = -\dfrac{b}{2a}$ and $k = f(h)$. The line of equation $x = h$ is the axis of symmetry of $\mathcal{P}$: for every real number $t$, the two points of $\mathcal{P}$ with $x$-coordinates $h - t$ and $h + t$ have the same $y$-coordinate. The point $S(h, k)$ where the axis meets $\mathcal{P}$ is its vertex: the lowest point of $\mathcal{P}$ if $a > 0$, the highest if $a < 0$.

Proof

The vertex form gives $f(x) = a(x - h)^2 + k$ with $h = -\frac{b}{2a}$ and $k = f(h)$. Then $$f(h + t) = at^2 + k = f(h - t):$$ the points with $x$-coordinates $h - t$ and $h + t$ are at the same height, symmetric about the line $x = h$. Moreover $f(x) - k = a(x - h)^2$ has the sign of $a$ and vanishes only at $x = h$: if $a > 0$, every other point of $\mathcal{P}$ is above $S$; if $a < 0$, below.

Method — Finding the Axis and the Vertex

For the parabola $y = ax^2 + bx + c$:

read $a$, $b$, $c$ and compute $h = -\dfrac{b}{2a}$: the axis is the line $x = h$;
compute $k = f(h)$: the vertex is $S(h, k)$;
check with two symmetric points: the parabola passes through $(0, c)$, so it must also pass through the mirror point $(2h, c)$, that is $f(2h) = c$. This is the horizontal line of the opening text, at the height $m = c$. (If $b = 0$, the axis is the $y$-axis itself and this check says nothing: use another point.)

Example

Determine the axis of symmetry and the vertex of the parabola $\mathcal{P}$ of equation $y = x^2 - 6x + 5$, then check with the point $(0, 5)$.

Answer

Here $a = 1$, $b = -6$, $c = 5$. $$\begin{aligned}[t] h &= -\frac{b}{2a} = -\frac{-6}{2 \times 1} = 3 && \scriptscriptstyle\text{(axis } x = 3\text{)} \\ k &= f(3) = 3^2 - 6 \times 3 + 5 = -4 && \scriptscriptstyle\text{(vertex } S(3, -4)\text{)} \end{aligned}$$ Since $a > 0$, $S$ is the lowest point of $\mathcal{P}$. Check: the mirror of $(0, 5)$ about the line $x = 3$ is $(6, 5)$, and $f(6) = 36 - 36 + 5 = 5$: both points are on $\mathcal{P}$.

Skills to practice

Finding the Axis of Symmetry and the Vertex

II.5 Key Features of a Graph with Technology

Method — Reading the Key Features of a Graph on the Screen

To study the graph of a function $f$ with a graphing calculator:

Enter $f(x)$ as the first function ($Y_1$) of the function editor.
Choose the window: the least and greatest values of $x$ and of $y$ shown on the screen. Start from the domain the question gives, or from a table of values of $f$, and adjust until every feature asked for is visible. A window that is too narrow hides an intercept or a turning point; a window that is too wide flattens the curve.
Read each feature with the analysis tools of the graph menu, never by eye:
- the $y$-intercept: the value of $f$ at $x = 0$;
- each $x$-intercept: the tool zero (or root);
- a local maximum or minimum: the tool maximum or minimum, which gives both coordinates of the turning point.
For the tools zero, maximum and minimum, bound the search: give a left bound and a right bound on either side of the feature, or a first guess close to it, so that the calculator finds that feature and not a neighbouring one.

The mock-up shows the maximum of $f(x) = -x^2 + x + 5$, in the window $-3 \le x \le 3$, $-5 \le y \le 10$.

Example

Let $f(x) = x^3 - 5x + 3$.

Choose a window that shows the intercepts and the turning points of the graph of $f$.
Find the $y$-intercept and the $x$-intercepts of the graph.
Find the local maximum point and the local minimum point.

Answer

A table of values of $f(x) = x^3 - 5x + 3$ at the integers guides the choice:

$x$ $-3$ $-2$ $-1$ $0$ $1$ $2$ $3$

$f(x)$ $-9$ $5$ $7$ $3$ $-1$ $1$ $15$

The sign of $f(x)$ changes between $-3$ and $-2$, between $0$ and $1$, and between $1$ and $2$: three $x$-intercepts, all between $-3$ and $3$. Since $f(-1) = 7$ is greater than both $f(-2) = 5$ and $f(0) = 3$, the curve has a highest point strictly between $-2$ and $0$: a local maximum lies there. Likewise $f(1) = -1$ is less than both $f(0) = 3$ and $f(2) = 1$: a local minimum lies between $0$ and $2$. The window $-3 \le x \le 3$, $-5 \le y \le 10$ shows them all; the two ends of the curve leave the screen, where there is nothing to read.
The $y$-intercept is $f(0) = 3$. The zero tool, bounded between $-3$ and $-2$, then between $0$ and $1$, then between $1$ and $2$, displays $-2.4908636$, $0.6566204$ and $1.8342432$: the $x$-intercepts are $x \approx -2.49$, $x \approx 0.657$ and $x \approx 1.83$. An equation of degree $3$ has at most three real roots, so none is missing.
The maximum tool, bounded between $-2$ and $0$, displays $X = -1.2909944$ and $Y = 7.3033148$; the minimum tool, bounded between $0$ and $2$, displays $X = 1.2909944$ and $Y = -1.3033148$. The local maximum point is $(-1.29, 7.30)$ and the local minimum point is $(1.29, -1.30)$.

Skills to practice

Finding Key Features with Technology

III The Discriminant and the Equation of a Parabola

III.1 Standard Form and the Discriminant

Recappreviously studied — click to reread

Proposition — Key Features from Standard Form

The graph of a quadratic function $f(x)=ax^2+bx+c$ is a parabola with:

Concavity: Determined by the sign of $a$.
- If $a > 0$, concave up.
- If $a < 0$, concave down.
Axis of Symmetry: The line $x = -\frac{b}{2a}$.
Vertex: The point $\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$.
y-intercept: The point $(0, c)$.

Proof

By expanding the vertex form $f(x) = a(x-h)^2+k$, we get:$$f(x) = a(x^2 - 2hx + h^2) + k = ax^2 - 2ahx + (ah^2+k)$$ Comparing this to the standard form $f(x)=ax^2+bx+c$, we can equate coefficients:$$b = -2ah \implies h = -\frac{b}{2a}$$ This confirms the formulas for the axis of symmetry and the vertex.

Definition — Discriminant

For a quadratic function $f(x) = ax^2 + bx + c$, the discriminant, denoted by $\boldsymbol{\Delta}$, is:$$ \boldsymbol{\Delta = b^2 - 4ac} $$

Proposition — Nature of Roots

If $\boldsymbol{\Delta > 0}$, there are two distinct x-intercepts. The roots are $x_1 = \frac{-b - \sqrt{\Delta}}{2a}$ and $x_2 = \frac{-b + \sqrt{\Delta}}{2a}$.
If $\boldsymbol{\Delta = 0}$, there is one x-intercept (at the vertex). The root is $x = \frac{-b}{2a}$.
If $\boldsymbol{\Delta < 0}$, there are no x-intercepts.

Proposition — Factorising a Quadratic Function

Let $f(x) = ax^2 + bx + c$ with $a \neq 0$ and $\Delta = b^2 - 4ac$.

If $\Delta > 0$, $f$ has two roots, denoted $x_1$ and $x_2$, and, for all $x$, $f(x) = a(x - x_1)(x - x_2)$.
If $\Delta = 0$, $f$ has one (double) root $x_0 = -\dfrac{b}{2a}$ and, for all $x$, $f(x) = a(x - x_0)^2$.
If $\Delta < 0$, $f$ has no root and cannot be written as a product of two first-degree factors.

Proof

By the vertex form, for all $x$, $$f(x) = a\left[\left(x + \frac{b}{2a}\right)^2 - \frac{\Delta}{4a^2}\right].$$

If $\Delta > 0$, then $\dfrac{\Delta}{4a^2} = \left(\dfrac{\sqrt{\Delta}}{2a}\right)^2$ and the bracket is a difference of two squares: $$f(x) = a\left(x + \frac{b}{2a} - \frac{\sqrt{\Delta}}{2a}\right)\left(x + \frac{b}{2a} + \frac{\sqrt{\Delta}}{2a}\right) = a\,(x - x_1)(x - x_2)$$ with $x_1 = \dfrac{-b - \sqrt{\Delta}}{2a}$ and $x_2 = \dfrac{-b + \sqrt{\Delta}}{2a}$, the two roots given by the quadratic formula.
If $\Delta = 0$, the bracket is $\left(x + \dfrac{b}{2a}\right)^2$, so $f(x) = a(x - x_0)^2$ with $x_0 = -\dfrac{b}{2a}$.
If $\Delta < 0$, then $-\dfrac{\Delta}{4a^2} > 0$, so the bracket is strictly positive for every $x$: $f(x)$ never vanishes. A product $a(x - p)(x - q)$ vanishes at $x = p$, so no such writing exists.

Example

Factorise, when possible: $f(x) = -3x^2 + 6x + 9$, $g(x) = 4x^2 - 4x + 1$ and $k(x) = x^2 + x + 1$.

Answer

$f$: $\Delta = 6^2 - 4 \times (-3) \times 9 = 36 + 108 = 144 > 0$, $\sqrt{\Delta} = 12$. The roots are $\dfrac{-6 - 12}{-6} = 3$ and $\dfrac{-6 + 12}{-6} = -1$, so $f(x) = -3(x + 1)(x - 3)$.
$g$: $\Delta = (-4)^2 - 4 \times 4 \times 1 = 0$, double root $x_0 = \dfrac{4}{8} = \dfrac12$, so $g(x) = 4\left(x - \dfrac12\right)^2 = (2x - 1)^2$.
$k$: $\Delta = 1 - 4 = -3 < 0$: $k$ has no root and cannot be factorised.

Method — Finding the Equation of a Parabola from Its Graph

A graph gives some key features of a parabola. The equation is written in the form in which these features appear, and one more point of the graph then gives the coefficient $a$.

Read the features given by the graph or by the statement: the $x$-intercepts, the vertex, the $y$-intercept, other points.
Choose the form that contains them:
- two $x$-intercepts $p$ and $q$: $y = a(x - p)(x - q)$, the factored form;
- the graph touches the $x$-axis at $p$: $y = a(x - p)^2$, since the two roots are equal and $(p, 0)$ is the vertex;
- the vertex $(h, k)$: $y = a(x - h)^2 + k$, the vertex form.
Find $a$: substitute the coordinates of another point of the graph and solve the equation obtained, whose unknown is $a$.
Check with a point not used yet, or with the direction of the parabola (upwards if $a > 0$, downwards if $a < 0$), and expand when the standard form is asked for.

When the graph shows neither the $x$-intercepts nor the vertex, the three coefficients of $y = ax^2 + bx + c$ are found from three points of the graph: each point gives one linear equation in $a$, $b$ and $c$, and the system of three equations is solved with technology (Method “Solving a Linear System with the Equation Solver”).

Example

Find, in the form $y = ax^2 + bx + c$, the equation of each parabola.

Graph (1) crosses the $x$-axis at $-1$ and $4$ and the $y$-axis at $8$; graph (2) has its vertex at $(1, -4)$ and passes through $(3, 4)$; graph (3) touches the $x$-axis at $2$ and crosses the $y$-axis at $12$.

Answer

The $x$-intercepts are $-1$ and $4$: the factored form is $y = a(x + 1)(x - 4)$. The $y$-intercept, the point $(0, 8)$, gives $a$: $$\begin{aligned}[t] a(0 + 1)(0 - 4) &= 8\\ -4a &= 8\\ a &= -2. \end{aligned}$$ Then $$\begin{aligned}[t] y &= -2(x + 1)(x - 4)\\ &= -2(x^2 - 3x - 4) && \text{(expand the product)}\\ &= -2x^2 + 6x + 8. && \text{(distribute $-2$)} \end{aligned}$$ Check: $a = -2 < 0$, and the parabola opens downwards, as on the graph.
The vertex is $(1, -4)$: the vertex form is $y = a(x - 1)^2 - 4$. The point $(3, 4)$ gives $a$: $$\begin{aligned}[t] a(3 - 1)^2 - 4 &= 4\\ 4a &= 8\\ a &= 2. \end{aligned}$$ Then $$\begin{aligned}[t] y &= 2(x - 1)^2 - 4\\ &= 2(x^2 - 2x + 1) - 4 && \text{(expand the square)}\\ &= 2x^2 - 4x - 2. \end{aligned}$$ Check with the point $(-1, 4)$, the mirror image of $(3, 4)$ about the axis of symmetry $x = 1$: $2 \times (-1)^2 - 4 \times (-1) - 2 = 4$.
The graph touches the $x$-axis at $2$: $y = a(x - 2)^2$. The $y$-intercept, the point $(0, 12)$, gives $a$: $$\begin{aligned}[t] a(0 - 2)^2 &= 12\\ 4a &= 12\\ a &= 3. \end{aligned}$$ Then $$\begin{aligned}[t] y &= 3(x - 2)^2\\ &= 3(x^2 - 4x + 4) && \text{(expand the square)}\\ &= 3x^2 - 12x + 12. \end{aligned}$$ Check: the constant term $12$ is the $y$-intercept read on the graph.

Skills to practice

Solving Parameter Problems with the Discriminant
Finding the Equation of a Parabola from Key Features
Finding the Equation of a Parabola from a Graph

IV The Intersection of Graphs

IV.1 Intersection of Curves

Recappreviously studied — click to reread

A common problem is to find where the graphs of two functions meet. The intersection points are the points $(x,y)$ that satisfy both equations simultaneously. By setting the function rules equal to each other, we can create a new equation to solve for the x-coordinates of these points.

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Method — Finding Intersection Points

To find the points of intersection between the graphs of two functions $f(x)$ and $g(x)$:

Set the functions equal to each other: $\boldsymbol{f(x) = g(x)}$.
Solve this equation for $x$. The solutions are the $x$-coordinates of the intersection points.
Substitute each $x$-coordinate back into either of the original functions ($f$ or $g$) to find the corresponding y-coordinate.

Example

Find the coordinates of the intersection points of the parabola $f(x) = x^2 - x - 2$ and the line $g(x) = x+1$.

Answer

We find the intersection points by setting $f(x) = g(x)$ and solving for $x$.$$ \begin{aligned}f(x) &= g(x) \\ x^2 - x - 2 &= x+1 \\ x^2 - 2x - 3 &= 0 \\ (x-3)(x+1) &= 0\end{aligned} $$ This gives two solutions for $x$: $x=3$ and $x=-1$.
Now we find the corresponding $y$-coordinates by substituting these values into the linear function $g(x)=x+1$:

If $x=3$, $y = 3+1 = 4$.
If $x=-1$, $y = -1+1 = 0$.

The graphs intersect at the points $\boldsymbol{(-1, 0)}$ and $\boldsymbol{(3, 4)}$.

Skills to practice

Finding Intersection Points Algebraically

IV.2 A Line and a Parabola

Proposition — A Line and a Parabola

Let $\mathcal{P}$ be the parabola $y = ax^2 + bx + c$, with $a \neq 0$, and $\mathcal{D}$ the line $y = mx + p$. The $x$-coordinates of their common points are the solutions of the quadratic equation $$ax^2 + (b - m)x + (c - p) = 0,$$ whose discriminant is $\Delta = (b - m)^2 - 4a(c - p)$.

If $\Delta > 0$, $\mathcal{D}$ and $\mathcal{P}$ have two common points: the line cuts the parabola.
If $\Delta = 0$, they have exactly one common point, and $\mathcal{P}$ lies on one side of $\mathcal{D}$: the line touches the parabola, and it is called a tangent to $\mathcal{P}$ at that point.
If $\Delta < 0$, they have no common point.

A vertical line $x = k$ always meets $\mathcal{P}$ at exactly one point, but $\mathcal{P}$ has points on both sides of it: a vertical line is not a tangent to a parabola.

Proof

A point $(x, y)$ lies on both $\mathcal{P}$ and $\mathcal{D}$ exactly when $y = mx + p$ and $ax^2 + bx + c = mx + p$, that is, when $y = mx + p$ and $g(x) = 0$, where $$g(x) = ax^2 + (b - m)x + (c - p).$$ Since $a \neq 0$, $g$ is a quadratic function, and its discriminant is $\Delta$. By the Proposition “Nature of Roots”, $g$ has two roots, one root or none according as $\Delta > 0$, $\Delta = 0$ or $\Delta < 0$; each root $x$ gives exactly one common point, $(x, mx + p)$.
If $\Delta = 0$, the Proposition “Factorising a Quadratic Function” gives $g(x) = a(x - x_0)^2$ for every $x$, with $x_0 = -\dfrac{b - m}{2a}$. For every $x \neq x_0$, the number $g(x) = (ax^2 + bx + c) - (mx + p)$ has the sign of $a$: the point of $\mathcal{P}$ with $x$-coordinate $x$ lies above the line if $a > 0$, below it if $a < 0$. So $\mathcal{P}$ lies on one side of $\mathcal{D}$ and meets it only at $x_0$.
For a vertical line $x = k$, the only common point is $(k, ak^2 + bk + c)$, and $\mathcal{P}$ has points with $x < k$ and points with $x > k$, on both sides of the line.

Example

Let $\mathcal{P}$ be the parabola $y = x^2 - 4x + 7$ and, for each real number $k$, let $\mathcal{D}_k$ be the line $y = 2x + k$.

Find the value of $k$ for which $\mathcal{D}_k$ is tangent to $\mathcal{P}$, and the point of contact.
Find the values of $k$ for which $\mathcal{D}_k$ cuts $\mathcal{P}$ at two points.

Answer

The $x$-coordinates of the common points are the solutions of $$\begin{aligned}[t] x^2 - 4x + 7 &= 2x + k\\ x^2 - 6x + (7 - k) &= 0, && \text{(subtract } 2x + k \text{ from both sides)} \end{aligned}$$ a quadratic equation with $a = 1$, $b = -6$ and $c = 7 - k$. Its discriminant is $$\begin{aligned}[t] \Delta &= (-6)^2 - 4 \times 1 \times (7 - k) && (\Delta = b^2 - 4ac)\\ &= 36 - 28 + 4k\\ &= 8 + 4k. \end{aligned}$$

By the Proposition “A Line and a Parabola”, $\mathcal{D}_k$ is tangent to $\mathcal{P}$ when $\Delta = 0$, that is $8 + 4k = 0$, so $k = -2$. The equation then reads $x^2 - 6x + 9 = 0$, that is $(x - 3)^2 = 0$: $x = 3$, and $y = 2 \times 3 - 2 = 4$. The point of contact is $(3, 4)$. Check on $\mathcal{P}$: $3^2 - 4 \times 3 + 7 = 4$.
$\mathcal{D}_k$ cuts $\mathcal{P}$ at two points when $\Delta > 0$, that is $8 + 4k > 0$, so $k > -2$. For $k < -2$ the line does not meet the parabola.

Example

Let $\mathcal{P}$ be the parabola $y = 0.6x^2 - 2.5x + 1.4$ and $\mathcal{D}$ the line $y = 0.5x - 1.1$.

Use technology to find the vertex of $\mathcal{P}$, to 3 significant figures, then check its $x$-coordinate with the axis of symmetry $x = -\dfrac{b}{2a}$.
Use technology to find the coordinates of the points of intersection of $\mathcal{P}$ and $\mathcal{D}$, to 3 significant figures.
Check with the discriminant that $\mathcal{P}$ and $\mathcal{D}$ have exactly two common points.

Answer

The parabola is entered as $Y_1 = 0.6x^2 - 2.5x + 1.4$ and graphed in the window $-1 \le x \le 6$, $-3 \le y \le 9$ (Method “Reading the Key Features of a Graph on the Screen”). Since $a = 0.6 > 0$, the vertex is the lowest point: the minimum tool, bounded between $1$ and $3$, displays $X = 2.0833333$ and $Y = -1.2041667$. The vertex is $(2.08, -1.20)$, the final zero being the third significant figure.
Check: $-\dfrac{b}{2a} = \dfrac{2.5}{1.2} = \dfrac{25}{12} = 2.0833\ldots$, the $x$-coordinate on the screen.
The line is entered as $Y_2 = 0.5x - 1.1$. The intersection tool (Method “Solving an Equation Graphically with Technology”), with a first guess close to each point, displays $X = 1.0566243$, $Y = -0.5716878$, then $X = 3.9433757$, $Y = 0.8716878$. The question asks for points, so both coordinates are kept: the points of intersection are $(1.06, -0.572)$ and $(3.94, 0.872)$.
The common points have $x$-coordinates the solutions of $$\begin{aligned}[t] 0.6x^2 - 2.5x + 1.4 &= 0.5x - 1.1\\ 0.6x^2 - 3x + 2.5 &= 0, && \text{(subtract } 0.5x - 1.1\text{)} \end{aligned}$$ whose discriminant is $\Delta = (-3)^2 - 4 \times 0.6 \times 2.5 = 9 - 6 = 3$. Since $\Delta > 0$, the Proposition “A Line and a Parabola” gives exactly two common points, as on the screen. The quadratic formula gives their exact $x$-coordinates, $\dfrac{3 \pm \sqrt{3}}{1.2}$, whose decimal values are the ones displayed.

Skills to practice

Counting the Common Points of a Line and a Parabola
Finding Intersection Points with Technology
Finding Tangent Lines with the Discriminant

V Problem Solving and Optimisation

V.1 Quadratic Optimisation

Recappreviously studied — click to reread

Because the graph of a quadratic function has a vertex that is either the absolute highest point (maximum) or lowest point (minimum) on the graph, quadratic functions are extremely useful for solving optimisation problems. These are problems where we want to find the maximum or minimum value of a certain quantity.

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Method — Solving Optimisation Problems

Read the problem to identify the quantity to be maximised or minimised.
Define variables and write a quadratic function that models this quantity.
Find the coordinates of the vertex $(h,k)$ of the parabola.
- $h = -\frac{b}{2a}$ gives the input value (e.g., time, length, price) where the optimum occurs.
- $k = f(h)$ gives the optimal (maximum or minimum) value itself.
Check that $h$ is an allowed value of the variable (for instance, a length or a time is positive).
Answer the question in the context of the problem, making sure to include units.

Example

A farmer wants to fence a rectangular area next to a river. He has 120 m of fencing and does not need to fence the side along the river. What dimensions maximise the area?

Answer

Define variables and the function: Let $x$ be the width of the rectangle (perpendicular to the river) and let $y$ be the length (parallel to the river). The total fencing is 120 m, which covers the two sides of width $x$ and the one side of length $y$. The constraint equation is: $$\begin{aligned} 2x + y &= 120 \\ y &= 120 - 2x\end{aligned} $$ The area, $A$, which we want to maximise, is given by the function: $$\begin{aligned}A(x) &= \text{width} \times \text{length}\\ &= x \cdot y \\ &= x(120 - 2x)\\ &= -2x^2 + 120x\end{aligned} $$
Analyse the function: This is a quadratic function with $a=-2$. Since $a<0$, the parabola opens downwards and has a maximum value at its vertex.
Find the vertex: The x-coordinate of the vertex gives the width that maximises the area. $$ x_{vertex} = -\frac{b}{2a} = -\frac{120}{2(-2)} = -\frac{120}{-4} = 30 $$ This means the optimal width is 30 m, an allowed value: the width lies between $0$ and $60$ m, so that $y = 120 - 2x$ stays positive.
Find the dimensions:
- Width: $x = 30$ m.
- Length: $y = 120 - 2x = 120 - 2(30) = 60$ m.
State the conclusion: The dimensions that maximise the area are 30 m by 60 m. The maximum area itself is $A(30) = 30 \times 60 = 1800 \text{ m}^2$.

Skills to practice

Solving Optimisation Problems

VI Quadratic Inequalities

VI.1 Solving Quadratic Inequalities

Recappreviously studied — click to reread

A quadratic inequality is an inequality that can be written in the form $ax^2+bx+c > 0$ (or $<, \leq, \geq$), with $a \neq 0$. Solving such an inequality involves finding the set of all $x$-values for which the statement is true. The key is to first find the roots of the corresponding equation, as these are the boundary points where the expression can change its sign.

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Proposition — Sign of a Quadratic Trinomial

Let $f(x) = ax^2 + bx + c$ with $a \neq 0$ and $\Delta = b^2 - 4ac$.

If $\Delta > 0$, with roots $x_1 < x_2$: $f(x)$ has the sign of $a$ outside the roots (for $x < x_1$ or $x > x_2$) and the opposite sign between the roots (for $x_1 < x < x_2$); $f(x_1) = f(x_2) = 0$.
If $\Delta = 0$, with double root $x_0$: $f(x)$ has the sign of $a$ for all $x \neq x_0$, and $f(x_0) = 0$.
If $\Delta < 0$: $f(x)$ has the sign of $a$ for all real $x$.

Proof

If $\Delta > 0$, the factored form gives $f(x) = a(x - x_1)(x - x_2)$. The factor $x - x_1$ is negative for $x < x_1$ and positive for $x > x_1$; likewise for $x - x_2$. A sign table of the product shows that $(x - x_1)(x - x_2)$ is positive outside $[x_1, x_2]$ and negative inside $(x_1, x_2)$; multiplying by $a$ gives the sign of $f(x)$.
If $\Delta = 0$, $f(x) = a(x - x_0)^2$ and $(x - x_0)^2 > 0$ for $x \neq x_0$.
If $\Delta < 0$, the vertex form $f(x) = a\left[\left(x + \dfrac{b}{2a}\right)^2 - \dfrac{\Delta}{4a^2}\right]$ has a strictly positive bracket for every $x$, so $f(x)$ has the sign of $a$.

Reading the sign on the parabola

The sign table matches the shape of the curve: a parabola with $a > 0$ is above the $x$-axis except between its two $x$-intercepts (if any), and a parabola with $a < 0$ is below the $x$-axis except between them. When $\Delta = 0$ the parabola only touches the $x$-axis at its vertex, and when $\Delta < 0$ it never meets the $x$-axis and stays entirely on one side.

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Example

Give the sign table of $f(x) = 2x^2 - 3x - 5$ and of $g(x) = -x^2 + 4x - 5$.

Answer

$f$: $a = 2 > 0$, $\Delta = 9 + 40 = 49 > 0$, $\sqrt{\Delta} = 7$, roots $x_1 = \dfrac{3 - 7}{4} = -1$ and $x_2 = \dfrac{3 + 7}{4} = \dfrac52$. $f(x)$ is positive outside the roots and negative between them.
$g$: $a = -1 < 0$, $\Delta = 16 - 20 = -4 < 0$: no root, $g(x)$ has the sign of $a$, so $g(x) < 0$ for every real $x$.

Method — Solving a Quadratic Inequality

To solve a quadratic inequality such as $ax^2+bx+c \ge 0$ when $\Delta> 0$:

Find the roots of the corresponding equation $ax^2+bx+c = 0$. Let the roots be $x_1$ and $x_2$. These are the “critical values” where the expression changes sign.
Construct a sign table with the roots placed in increasing order on the number line.
Determine the sign pattern. Instead of testing each factor, we can use the properties of the parabola $y=ax^2+bx+c$:
- If $a>0$, the parabola opens upwards , so the expression is positive outside the roots and negative between them.
- If $a<0$, the parabola opens downwards , so the expression is negative outside the roots and positive between them.
State the solution set by choosing the interval(s) that satisfy the inequality.

If $\Delta = 0$ or $\Delta < 0$, there are no two roots to split the number line: $ax^2+bx+c$ keeps the sign of $a$ for every $x$ (except at the double root, where it is $0$), and the solution set is read directly.

Example

Solve the inequality $x^2 - x - 6 < 0$.

Answer

Find the roots: First, we solve the corresponding equation $x^2 - x - 6 = 0$. By factorising, we get $(x-3)(x+2) = 0$. The roots are $x=3$ and $x=-2$.
Construct the sign table: The critical values are $-2$ and $3$. The quadratic expression is $x^2 - x - 6$. Since the leading coefficient is $a=1$ (positive), the parabola opens upwards. This means the expression is positive outside the roots and negative between them.
State the solution: The inequality is $x^2 - x - 6 < 0$, so we are looking for the interval where the expression is negative. According to the sign table, this is the interval between $-2$ and $3$. Since the inequality is strict ($<$), the endpoints are not included.

The solution is $-2 < x < 3$.

Skills to practice

Solving Quadratic Inequalities
Solving Inequalities by Graphing

VII Going further

Skills to practice

Reasoning about Quadratic Functions
Solving Inequalities and Parameter Problems
Solving Problems without Technology
Solving Problems with Technology
Quiz

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\(x\)	\(-3\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(3\)
\(f(x)\)	\(-9\)	\(5\)	\(7\)	\(3\)	\(-1\)	\(1\)	\(15\)