International Baccalaureate · MYP 4

Quadratic Functions

⌚ ~186 min ▢ 21 blocks ✓ 95 exercises ➣ Prerequisites : Quadratic Equations, Coordinate Geometry and Straight Lines, Radicals and Surds, Sets, Number Sets and Intervals

I Quadratic Functions and Their Graphs

I.1 The Square Function

Definition — Square Function

The square function is defined on $\mathbb{R}$ by $f(x) = x^2$. This means that each input value $x$ is multiplied by itself to give the output.

Domain: All real numbers ($\mathbb{R}$).
Shape: A curve called a parabola opening upwards.

Proposition — Properties

For any real number $x$, $x^2 \geq 0$.
The square function is strictly decreasing on $(-\infty, 0]$ and strictly increasing on $[0, +\infty)$.
The square function is even: its graph is symmetric with respect to the $y$-axis ($f(-x) = f(x)$).

Proof

Let $f(x) = x^2$.

The product of two real numbers with the same sign is always non-negative, so $x^2 = x \times x \geq 0$.
Parity: $f(-x) = (-x)^2 = (-x) \times (-x) = x^2 = f(x)$. Thus, $f$ is even.
Variations:
- Let $a$ and $b$ be two real numbers such that $0 \leq a < b$.$$f(b) - f(a) = b^2 - a^2 = (b - a)(b + a)$$ Since $b > a$, then $(b - a) > 0$. Since $a$ and $b$ are non-negative and not both zero, $(b + a) > 0$.
  The product is positive, so $f(b) - f(a) > 0$, meaning $f(a) < f(b)$. $f$ is strictly increasing on $[0, +\infty)$.
- Let $a$ and $b$ be two real numbers such that $a < b \leq 0$.
  In this case, $(b - a) > 0$ but $(b + a) < 0$.
  The product is negative, so $f(b) - f(a) < 0$, meaning $f(a) > f(b)$. $f$ is strictly decreasing on $(-\infty, 0]$.

Example

The area of a square with side length $x$ is given by $A(x) = x^2$.

If $x = 4$ m, then $A(4) = 4^2 = 16$ m$^2$.

Skills to practice

Comparing Squares

I.2 Quadratic Functions

Linear functions have straight lines as graphs. Quadratic functions come next: functions containing an $x^2$ term, whose graphs are symmetrical curves called parabolas. In mathematics and science, quadratic functions model, for example, the path of a thrown ball or the shape of a satellite dish. We learn to identify and graph these functions, and to solve the equations that come with them.

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A quadratic function is a polynomial of degree two. Its defining feature is the $x^2$ term, which distinguishes it from a linear function.

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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 $\boldsymbol{a \neq 0}$.

Example

For $f(x) = x^2 - 3x + 1$, evaluate $f(2)$.

Answer

$\begin{aligned}[t]f(2) &= (2)^2 - 3(2) + 1 \\&= 4 - 6 + 1 \\&= -1\end{aligned}$

Skills to practice

Recognising Quadratic Functions
Identifying the Coefficients
Finding Images
Justifying That a Function Is Quadratic

I.3 The Graph of a Quadratic Function

The graph of every quadratic function has a characteristic U-shape (either upright or inverted) called a parabola. This shape is found in many areas of science and nature.
The parabola is one of the conic sections, which are the group of curves obtained by intersecting a cone with a plane. A parabola is produced by intersecting the cone with a plane parallel to one of its generating lines and not passing through its apex.

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Example

Sketch the graph of $f(x) = x^2 - x - 1$.

Answer

A table of values is

$x$	$-2$	$-1$	0	1	2
$y$	5	1	$-1$	$-1$	1

The coefficient $a$ in $ax^2+bx+c$ controls whether the parabola opens upwards or downwards. To see this interactively, use the GeoGebra animation at https://www.geogebra.org/m/gn3c2sqe.

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Proposition — The Parabola and Its Axis of Symmetry

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

Axis of symmetry: the parabola has a vertical axis of symmetry. Two points of the parabola with the same $y$-value are symmetric about this axis: if $f(u) = f(v)$ with $u \neq v$, the axis is the line $x = \dfrac{u+v}{2}$, through the midpoint of their $x$-coordinates.
Vertex: the point of the parabola that lies on the axis. It is the lowest point if the parabola opens upwards, and the highest point if it opens downwards.
$y$-intercept: the number $f(0) = c$; the parabola crosses the $y$-axis at the point $(0, c)$.
$x$-intercepts: the numbers $x$ such that $f(x) = 0$, also called the roots of $f$; they are the $x$-coordinates of the points where the parabola meets the $x$-axis. Two such points have the same $y$-coordinate $0$, so, for two roots $x_1$ and $x_2$, the axis is $x = \dfrac{x_1+x_2}{2}$.

Proposition — Direction of Opening

For any quadratic function $f(x) = ax^2 + bx + c$:

If $a > 0$, the parabola opens upwards (it is concave up) and has a minimum value at its vertex. .
If $a < 0$, the parabola opens downwards (it is concave down) and has a maximum value at its vertex. .

Skills to practice

Determining the Direction of Opening from the Sign of $a$
Plotting and Reading Parabolas

I.4 Finding $x$ from $y$

Solving the equation $f(x) = k$, for a quadratic function $f$ and a given number $k$, means finding the input value(s) $x$ that give the output $k$. This process results in a quadratic equation. A particularly important case is solving $f(x) = 0$, which gives the $x$-intercepts of the parabola.

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Example

Let $f(x) = x^2 - 2x - 3$. Find the values of $x$ such that $f(x) = 5$.

Answer

We solve the equation $f(x) = 5$: $$\begin{aligned} x^2 - 2x - 3 &= 5\\ x^2 - 2x - 8 &= 0 &&\text{(subtract $5$ from both sides)}\\ (x - 4)(x + 2) &= 0 &&\text{(factorise: $(-4) \times 2 = -8$ and $(-4) + 2 = -2$)} \end{aligned}$$ By the null factor law, $x - 4 = 0$ or $x + 2 = 0$, so $x = 4$ or $x = -2$.
Check: $f(4) = 16 - 8 - 3 = 5$ and $f(-2) = 4 + 4 - 3 = 5$.
The values of $x$ such that $f(x) = 5$ are $\boldsymbol{-2}$ and $\boldsymbol{4}$.

Skills to practice

Finding the Inputs That Give an Output
Calculating the Inputs That Give an Output

II Transformations of Graphs

II.1 Translations

Function transformations allow us to take a basic “parent” function and modify it to create a new, related function. By applying a sequence of transformations, we can shift, stretch, compress, or reflect the graph of the parent function.
Understanding these transformations is essential, as it allows us to predict the graph of a complex function based on a simpler one and to model real-world phenomena by adjusting a basic function to fit observed data. The transformations are vertical and horizontal translations, stretches and compressions, and reflections.

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Definition — Vertical Translation by Adding a Constant

A vertical translation shifts the graph of a function up or down. The transformation is defined by:$$g(x) = f(x) + k$$ This transformation maps a point $(x, y)$ on the graph of $f$ to a new point $(x, y+k)$ on the graph of $g$.

If $k>0$, the graph is translated $\boldsymbol{k}$ units up.
If $k<0$, the graph is translated $\boldsymbol{|k|}$ units down.

Definition — Horizontal Translation

A horizontal translation shifts the graph of a function left or right. The transformation is defined by:$$g(x) = f(x-h)$$ This transformation maps a point $(x, y)$ on the graph of $f$ to a new point $(x+h, y)$ on the graph of $g$.

If $h>0$, the graph is translated $\boldsymbol{h}$ units to the right.
If $h<0$, the graph is translated $\boldsymbol{|h|}$ units to the left.

Example

From the graph of $f(x) = x^2$, plot $g(x) = f(x) + 2$ and $h(x) = f(x - 3)$.

$g(x) = f(x) + 2$: every output is raised by $2$ — a vertical translation $2$ units up. The vertex $(0, 0)$ moves to $(0, 2)$.
$h(x) = f(x - 3)$: the input $x = 3$ now plays the role $0$ used to play — a horizontal translation $3$ units to the right. The vertex $(0,0)$ moves to $(3, 0)$.

Translation by a Vector

The vertical and the horizontal translations combine into one. The translation by the vector $\begin{pmatrix} h \\ k \end{pmatrix}$ moves each point $(x, y)$ of the graph of $f$ to the point $(X, Y) = (x + h, y + k)$. Since $x = X - h$ and $Y = f(x) + k$, the image points satisfy $Y = f(X - h) + k$; writing $(x, y)$ again for the coordinates, the image of the graph of $y = f(x)$ is the graph of $$y = f(x - h) + k.$$ For example, the translation by the vector $\begin{pmatrix} 3 \\ -2 \end{pmatrix}$ maps the graph of $y = x^2$ onto the graph of $y = (x - 3)^2 - 2$, and its vertex $(0, 0)$ to $(3, -2)$.

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Skills to practice

Translating Graphs

II.2 Stretches

Definition — Vertical Stretch

A vertical stretch stretches or compresses the graph of a function vertically. It is defined, for a number $a \neq 0$, by:$$g(x) = a \cdot f(x)$$ This transformation maps a point $(x, y)$ to $(x, ay)$.

If $|a|>1$, the graph is stretched vertically by a factor of $|a|$.
If $0 < |a| < 1$, the graph is compressed vertically by a factor of $|a|$.
If $a < 0$, the stretch or compression is followed by a reflection in the $x$-axis ($a = -1$ gives the reflection alone).

Definition — Horizontal Stretch

A horizontal stretch stretches or compresses the graph of a function horizontally. It is defined, for a number $b \neq 0$, by:$$g(x) = f(bx)$$ This transformation maps a point $(x, y)$ to a new point $(\frac{x}{b}, y)$.

If $|b|>1$, the graph is compressed horizontally by a factor of $\frac{1}{|b|}$.
If $0 < |b| < 1$, the graph is stretched horizontally by a factor of $\frac{1}{|b|}$.
If $b < 0$, the stretch or compression is followed by a reflection in the $y$-axis ($b = -1$ gives the reflection alone).

Example

From the graph of $f(x) = x^2$, plot $a(x) = 2f(x)$ and $b(x) = f(2x)$.

$a(x) = 2f(x) = 2x^2$: every output is doubled — a vertical stretch by factor $2$. The point $(1, 1)$ moves to $(1, 2)$.
$b(x) = f(2x) = 4x^2$: the input is consumed twice as fast — a horizontal compression with scale factor $\frac12$: every $x$-coordinate is halved. The point $(1, 1)$ moves to $\left(\frac12, 1\right)$.

The two results differ: doubling the output is not the same as doubling the input.

Stretches and Scale Factors

The number by which a stretch multiplies the distances to the axis that stays fixed is its scale factor. For $a > 0$, the graph of $y = a f(x)$ is the image of the graph of $y = f(x)$ under the vertical stretch with scale factor $a$: each point $(x, y)$ goes to $(x, ay)$. For $b > 0$, the graph of $y = f(bx)$ is its image under the horizontal stretch with scale factor $\dfrac{1}{b}$: each point $(x, y)$ goes to $\left(\dfrac{x}{b}, y\right)$. A scale factor greater than $1$ moves the points away from the axis that stays fixed (the $x$-axis for a vertical stretch, the $y$-axis for a horizontal one), a scale factor between $0$ and $1$ brings them closer to it: the graph of $y = f(2x)$ is the image of the graph of $f$ under the horizontal stretch with scale factor $\dfrac{1}{2}$, a compression.

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Skills to practice

Stretching Graphs

II.3 Reflections

Definition — Reflection in the x-axis

A reflection in the x-axis flips the graph of a function vertically. It is a special case of vertical stretch where $a=-1$. It is defined by:$$g(x) = -f(x)$$ This transformation maps a point $(x, y)$ to $(x, -y)$.

Definition — Reflection in the y-axis

A reflection in the y-axis flips the graph of a function horizontally. It is a special case of horizontal stretch where $b=-1$. It is defined by:$$g(x) = f(-x)$$ This transformation maps a point $(x, y)$ to $(-x, y)$.

Example

From the graph of $f(x) = \sqrt{x}$, plot $r(x) = -f(x)$ and $s(x) = f(-x)$.

$r(x) = -\sqrt{x}$: every output changes sign — a reflection in the $x$-axis. The curve flips below the axis; the point $(4, 2)$ moves to $(4, -2)$.
$s(x) = \sqrt{-x}$: every input changes sign — a reflection in the $y$-axis. The curve flips to the left half-plane (defined for $x \le 0$); the point $(4, 2)$ moves to $(-4, 2)$.

Skills to practice

Reflecting Graphs

III Forms of a Quadratic Function

III.1 Vertex Form

A quadratic function can be expressed in three main forms. Each form highlights different key features of its graph, the parabola. Being able to recognise and convert between these forms is a fundamental skill.

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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 — 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 $x=h$.

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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Method — Converting from Vertex to Standard Form

To convert from vertex to standard form, expand the squared term and collect like terms.

Example

Convert $f(x) = 2(x-3)^2 + 1$ to standard form.

Answer

$$ \begin{aligned}f(x) &= 2(x^2 - 6x + 9) + 1 &&\color{gray}{(\text{Expand }(x-3)^2}) \\ &= 2x^2 - 12x + 18 + 1 &&\color{gray}{(\text{Distribute the 2})} \\ &= \boldsymbol{2x^2 - 12x + 19} &&\color{gray}{(\text{Combine constants})}\end{aligned} $$

Skills to practice

Converting from Vertex Form to Standard Form
Finding the Vertex by Completing the Square
Completing the Square in a Quadratic Expression

III.2 Factored Form

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 $\boldsymbol{a \neq 0}$.

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 $\frac{p+q}{2}$.

Method — Converting from Factored to Standard Form

To convert from factored form to standard form, expand the brackets and collect like terms.

Example

Convert $f(x) = 2(x-3)(x-1)$ to standard form.

Answer

$$ \begin{aligned}f(x) &=2(x-3)(x-1)\\ &= 2(x^2 - x - 3x + 3) \\ &= 2(x^2 - 4x + 3) \\ &= \boldsymbol{2x^2 - 8x + 6}\end{aligned} $$

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

Method — Checking a Root and Factorising with a Known Root

Let $f(x) = ax^2 + bx + c$ be a quadratic function. A root read on a graph, found in a table of values or simply guessed is only a conjecture until it is checked.

Check the root: $x_1$ is a root of $f$ when $f(x_1) = 0$. Substitute and compute.
Write the factored form with the second root unknown: a quadratic function with a root $x_1$ can be written $f(x) = a(x - x_1)(x - x_2)$, with the same $a$ as in the standard form (accepted without proof; the second root $x_2$ may be equal to $x_1$; the Definition above writes the roots $p$ and $q$).
Find $x_2$ with the constant term: expanding $a(x - x_1)(x - x_2)$ gives the constant term $a x_1 x_2$, which must be equal to $c$. Solve $a x_1 x_2 = c$.
Check by expanding the factored form.

If the known root is $x_1 = 0$, then $c = 0$ and $x$ is a common factor: $f(x) = x(ax + b)$.

Example

Let $f(x) = 2x^2 - 2x - 12$. Its graph seems to cross the $x$-axis at $x = 3$.

Check that $3$ is a root of $f$.
Factorise $f(x)$.
Give the two roots of $f$.

Answer

$\begin{aligned}[t]f(3) &= 2 \times 3^2 - 2 \times 3 - 12 \\ &= 18 - 6 - 12 \\ &= 0\end{aligned}$
So $3$ is a root of $f$.
Since $3$ is a root and $a = 2$, $f(x) = 2(x - 3)(x - x_2)$. The constant term of this product is $2 \times 3 \times x_2 = 6x_2$, and the constant term of $f$ is $-12$: $$\begin{aligned}6x_2 &= -12 \\ x_2 &= -2\end{aligned}$$ So $f(x) = 2(x - 3)(x + 2)$. Check: $$\begin{aligned}2(x - 3)(x + 2) &= 2(x^2 + 2x - 3x - 6) \\ &= 2(x^2 - x - 6) \\ &= 2x^2 - 2x - 12\end{aligned}$$
A product is zero when one of its factors is zero: the roots of $f$ are $3$ and $-2$.

Skills to practice

Converting from Factored Form to Standard Form
Finding the Second Root from a Known Root
Checking a Root and Factorising

III.3 Key Features 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 $f(x) = -0.5x^2 + 3x + 2$.

Choose a window that shows the intercepts and the turning point of the graph of $f$.
Find the $y$-intercept, and the $x$-intercepts to 3 significant figures.
Find the maximum point of the graph.
Check that the two $x$-intercepts are symmetric about the vertical line through the maximum point.

Answer

A table of values of $f(x) = -0.5x^2 + 3x + 2$ at the integers guides the choice:

$x$ $-2$ $-1$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$

$f(x)$ $-6$ $-1.5$ $2$ $4.5$ $6$ $6.5$ $6$ $4.5$ $2$ $-1.5$ $-6$

The sign of $f(x)$ changes between $-1$ and $0$ and between $6$ and $7$: two $x$-intercepts. The values rise up to $f(3) = 6.5$, then fall back symmetrically: $f(x) = ax^2 + bx + c$ with $a = -0.5 < 0$, so the graph is a parabola opening downwards, with its highest point near $x = 3$. The window $-2 \le x \le 8$, $-2 \le y \le 7$ shows the intercepts and the turning point; the two ends of the curve leave the screen, where there is nothing to read.
The $y$-intercept is $f(0) = 2$. The zero tool, bounded between $-1$ and $0$, then between $6$ and $7$, displays $X = -0.6055513$ and $X = 6.6055513$: the $x$-intercepts are $x \approx -0.606$ and $x \approx 6.61$. A parabola has at most two $x$-intercepts, so none is missing.
The maximum tool, bounded between $2$ and $4$, displays $X = 3$ and $Y = 6.5$; some models display a value such as $X = 2.9999996$, because the tool searches the maximum numerically and stops when it is close enough to it. The reading is confirmed exactly by completing the square: $$\begin{aligned}[t] f(x) &= -0.5\left(x^2 - 6x\right) + 2\\ &= -0.5\left((x - 3)^2 - 9\right) + 2 && (x^2 - 6x = (x - 3)^2 - 9)\\ &= -0.5(x - 3)^2 + 6.5. \end{aligned}$$ Since $-0.5(x - 3)^2 \le 0$, $f(x) \le 6.5$, with equality only for $x = 3$: the maximum point is exactly $(3, 6.5)$, as the table, symmetric about $x = 3$, suggested.
The two $x$-intercepts lie at the same distance from the vertical line $x = 3$: $$\begin{aligned}[t] 3 - (-0.6055513) &= 3.6055513, \\ 6.6055513 - 3 &= 3.6055513. \end{aligned}$$ The readings are consistent with the symmetry of the parabola about the line $x = 3$.

Key Features of a Parabola on the Screen

For a quadratic function $f(x) = ax^2 + bx + c$, with $a \neq 0$, the three steps of the method give every key feature: a parabola has no asymptote. The graph of such a function has exactly one $y$-intercept, $f(0) = c$, at most two $x$-intercepts, and exactly one turning point, the vertex: a maximum point when $a < 0$, a minimum point when $a > 0$. The vertical line through the vertex is an axis of symmetry of the parabola, so the two $x$-intercepts, when there are two, lie at the same distance from it. Compared with their full displayed values, the two distances are equal, up to the rounding of the screen; two readings that are clearly not symmetric about this line reveal a reading error.

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Skills to practice

Reading the Key Features of a Parabola

IV Going further

Skills to practice

Solving Projectile Problems
Investigating the Graphs of $y \equal a(x-h)^2+k$
Quiz

\(x\)	\(-2\)	\(-1\)	0	1	2
\(y\)	5	1	\(-1\)	\(-1\)	1

\(x\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)	\(6\)	\(7\)	\(8\)
\(f(x)\)	\(-6\)	\(-1.5\)	\(2\)	\(4.5\)	\(6\)	\(6.5\)	\(6\)	\(4.5\)	\(2\)	\(-1.5\)	\(-6\)