Quadratic functions

Introduction

This page is about quadratic functions and their graphs. We learn what a quadratic function is, how its graph (a parabola) looks, and how to find the vertex, the axis of symmetry, the intercepts, and the maximum or minimum value.

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 constants and $a\ne 0$. The highest power of the variable is $2$.

For example, the following are quadratic functions:

$f(x)=2x^2$, $g(x)=4x^2-5$, $h(t)=t^2+7t-6$

Graph of a quadratic function

The graph of a quadratic function is a curve called a parabola. If $a\gt 0$, the parabola opens upward; if $a\lt 0$, it opens downward.

Vertex of a parabola

The vertex is the turning point of the parabola — the lowest point if it opens upward, or the highest point if it opens downward. The vertex is written as the ordered pair $(h,k)$.

Axis of symmetry

The axis of symmetry is the vertical line that passes through the vertex and divides the parabola into two mirror-image halves. Its equation is

$x=h$

Standard form of a quadratic function

The standard form (also called the vertex form) of a quadratic function is

$f(x)=a(x-h)^2+k$

In this form, the vertex is $(h,k)$ and the axis of symmetry is $x=h$.

Shapes of parabolas

The number $a$ controls the direction and the width of the parabola. The sign of $a$ tells the direction: $a\gt 0$ opens upward and $a\lt 0$ opens downward. The size of $|a|$ tells the width, compared with the graph of $f(x)=x^2$:

If $|a|\gt 1$, the parabola is narrower than $f(x)=x^2$. If $|a|\lt 1$, it is wider.

From narrowest to widest: $y=2x^2$ (red), $y=x^2$ (blue), and $y=\dfrac{1}{2}x^2$ (green).

Example: For each quadratic function, tell whether the graph opens up or down and whether it is wider, narrower, or the same shape as the graph of $f(x)=x^2$.

(a) $f(x)=\dfrac{7}{9}x^2$ (b) $f(x)=-4(x+2)^2+5$ (c) $f(x)=x^2-8$

Solution:

(a) Here $a=\dfrac{7}{9}$. It is positive, so the parabola opens up. Since $\dfrac{7}{9}\lt 1$, the parabola is wider than the graph of $f(x)=x^2$.

(b) Here $a=-4$. It is negative, so the parabola opens down. Since $|-4|=4\gt 1$, the parabola is narrower than the graph of $f(x)=x^2$.

(c) Here $a=1$. It is positive, so the parabola opens up. Since $|1|=1$, the parabola has the same shape as the graph of $f(x)=x^2$.

Finding the vertex from standard form

When a quadratic function is written in standard form $f(x)=a(x-h)^2+k$, the vertex is read off directly as $(h,k)$.

Example: Find the vertex of each parabola.

(a) $f(x)=(x-6)^2-4$ (b) $f(x)=7x^2$ (c) $f(x)=-2(x+11)^2+5$

Solution:

(a) Comparing with $f(x)=a(x-h)^2+k$, we have $h=6$ and $k=-4$. The vertex is $(6,-4)$.

(b) Write $7x^2=7(x-0)^2+0$, so $h=0$ and $k=0$. The vertex is $(0,0)$.

Finding the vertex from the general form

If a quadratic function is given in the general form $f(x)=ax^2+bx+c$, we can find the vertex $(h,k)$ using the formulas

$h=-\dfrac{b}{2a}$ and $k=f(h)$

The axis of symmetry is then $x=h$.

Example 1: Find the vertex and the axis of symmetry of the parabola $f(x)=x^2-6x+8$.

Solution: Here $a=1$, $b=-6$, and $c=8$. Find $h$:

$h=-\dfrac{b}{2a}=-\dfrac{-6}{2\cdot 1}=3$

Find $k$ by evaluating $f$ at $x=3$:

$k=f(3)=3^2-6\cdot 3+8=9-18+8=-1$

So the vertex is $(3,-1)$, and the axis of symmetry is $x=3$.

Example 2: Find the vertex and the axis of symmetry of the parabola $f(x)=5x^2+30x-47$.

Solution: Here $a=5$, $b=30$, and $c=-47$. Find $h$:

$h=-\dfrac{b}{2a}=-\dfrac{30}{2\cdot 5}=-3$

Find $k$ by evaluating $f$ at $x=-3$:

$k=f(-3)=5(-3)^2+30(-3)-47=45-90-47=-92$

So the vertex is $(-3,-92)$, and the axis of symmetry is $x=-3$.

Parabola with a horizontal axis of symmetry

So far, every parabola has had a vertical axis of symmetry and has come from an equation of the form $y=ax^2+bx+c$. A parabola can also have a horizontal axis of symmetry. Its equation has the form

$x=ay^2+by+c$

Such a parabola opens to the right if $a\gt 0$ and to the left if $a\lt 0$.

x- and y-intercepts of a parabola

The x-intercepts are the points where the parabola crosses the $x$-axis (where $y=0$). The y-intercept is the point where it crosses the $y$-axis (where $x=0$).

The discriminant and the number of x-intercepts

A parabola with a vertical axis of symmetry can have two, one, or no $x$-intercepts.

We can find the number of $x$-intercepts from the discriminant, $b^2-4ac$:

Discriminant	Number of x-intercepts
positive	$2$
zero	$1$
negative	$0$

Example 1: The parabola $f(x)=6x^2-9x+2$ has a vertical axis of symmetry. Find the discriminant and use it to determine the number of $x$-intercepts.

Solution: Here $a=6$, $b=-9$, and $c=2$. Find the discriminant:

$b^2-4ac=(-9)^2-4\cdot 6\cdot 2=81-48=33$

The discriminant is positive, so the parabola has two $x$-intercepts.

Example 2: The parabola $f(x)=x^2+2x+12$ has a vertical axis of symmetry. Find the discriminant and use it to determine the number of $x$-intercepts.

Solution: Here $a=1$, $b=2$, and $c=12$. Find the discriminant:

$b^2-4ac=2^2-4\cdot 1\cdot 12=4-48=-44$

The discriminant is negative, so the parabola has no $x$-intercept.

Maximum or minimum of a quadratic function

The vertex gives the maximum or minimum value of a quadratic function. If the parabola opens upward ($a\gt 0$), the vertex is the lowest point, so the function has a minimum value. If it opens downward ($a\lt 0$), the vertex is the highest point, so the function has a maximum value. In both cases the maximum or minimum value is $k$, and it occurs at $x=h$.

Example 1: A ball is thrown from an initial height of $3$ feet with an initial velocity of $160$ ft/s. The height $h$, in feet, after $t$ seconds is given by

$h(t)=-16t^2+160t+3$

How long does it take the ball to reach its maximum height, and what is the maximum height?

Solution: Here $a=-16$, $b=160$, and $c=3$. Since $a\lt 0$, the parabola opens downward, so the function has a maximum at the vertex. The time at which it occurs is

$t=-\dfrac{b}{2a}=-\dfrac{160}{2(-16)}=5$

So the ball reaches its maximum height after $5$ seconds. The maximum height is the function value at $t=5$:

$h(5)=-16(5)^2+160(5)+3=-400+800+3=403$

The maximum height is $403$ feet.

Example 2: An object is thrown upward with an initial velocity of $224$ ft/s. The height $h$, in feet, after $t$ seconds is given by

$h(t)=224t-16t^2$

(a) Find the maximum height reached by the object. (b) Find the time it takes the object to hit the ground.

Solution: Write the function as $h(t)=-16t^2+224t$, so $a=-16$, $b=224$, and $c=0$.

(a) Since $a\lt 0$, the function has a maximum at the vertex. It occurs at

$t=-\dfrac{b}{2a}=-\dfrac{224}{2(-16)}=7$

The maximum height is

$h(7)=-16(7)^2+224(7)=-784+1568=784$

The maximum height is $784$ feet.

(b) The object hits the ground when its height is $0$. Set $h(t)=0$ and solve:

$224t-16t^2=0$

Factor out $16t$:

$16t(14-t)=0$

So $t=0$ or $14-t=0$, which gives $t=14$. The time $t=0$ is the start, so the object hits the ground after $14$ seconds.

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