International Baccalaureate · MYP 5

Quadratic Functions

⌚ ~138 min ▢ 15 blocks ✓ 66 exercises ➣ Prerequisites : Quadratic Equations, Functions, Quadratic Functions

I Converting to Vertex Form

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 $k' = k$.

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 $\boldsymbol{f(x) = 2(x-3)^2 + 1}$.

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: $-2(x - 2)^2 + 5 = -2(x^2 - 4x + 4) + 5 = -2x^2 + 8x - 3$.

Skills to practice

Converting to Vertex Form

II Finding the Axes Intercepts

Method — Solving $f(x)\equal k$

To solve $f(x) = k$ for a quadratic function $f(x) = ax^2 + bx + c$ and a given number $k$, first bring the equation to the form $ax^2 + bx + (c - k) = 0$. Since it is quadratic, there may be 0, 1 or 2 real solutions for $x$, which can be found using the quadratic formula $x = \dfrac{-b \pm \sqrt{\Delta}}{2a}$, where $\Delta = b^2 - 4a(c - k)$ is the number under the square root.

Example

For $f(x) = 2x^2 - 5x + 2$, find the $x$-intercepts of the function.

Answer

To find the x-intercepts, we set $f(x) = 0$: $2x^2 - 5x + 2 = 0$. We identify $a=2$, $b=-5$, $c=2$.

First, find the discriminant: $\begin{aligned}[t]\Delta &= b^2 - 4ac \\&= (-5)^2 - 4(2)(2) \\&= 25 - 16 \\&= 9\end{aligned}$
As $\Delta > 0$, there are 2 distinct real roots.
Use the quadratic formula: $\begin{aligned}[t]x &= \frac{-b \pm \sqrt{\Delta}}{2a} \\x &= \frac{-(-5) \pm \sqrt{9}}{2(2)} \\x &= \frac{5 \pm 3}{4}\end{aligned}$
The two roots are: $x_1 = \frac{5 - 3}{4} = \frac{2}{4} = \frac{1}{2}$ and $x_2 = \frac{5 + 3}{4} = \frac{8}{4} = 2$.

The $x$-intercepts are at $\boldsymbol{x = \frac{1}{2}}$ and $\boldsymbol{x = 2}$.

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 $(\frac{p+q}{2},f(\frac{p+q}{2}))$.

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.

Skills to practice

Solving $f(x) \equal k$ with the Formula
Writing the Factored Form from the Roots

III Finding the Axis of Symmetry and the Vertex

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 $\mathcal{P}$.

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 $(0, c)$.

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

Finding the Axis of Symmetry and the Vertex
Finding and Classifying the Vertex
Finding a Coefficient from the Axis of Symmetry
Drawing the Variation Table

IV Finding the Equation of a Parabola

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

Choosing the Equation from the Vertex
Choosing the Equation from the x-Intercepts
Finding the Equation from Key Features
Finding the Equation from a Graph

V Solving Optimisation Problems

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

Modelling and Optimising

VI Going further

Skills to practice

Reasoning about Parabolas
Solving Problems with Quadratic Functions
Investigating Patterns in the Vertices of Parabolas
Quiz