# Quadratic Functions in Standard Form

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

and the properties of their graphs such as vertex and x and y intercepts are explored, interactively, using an applet.

The graph of a quadratic function is "U" shaped and is called a parabola. The exploration is carried by changing the values of all 3 coefficients $a$, $h$ and $k$. Once you finish the present tutorial, you may want to go through another tutorial on graphing quadratic functions.

This form of the quadratic function is also called the vertex form.

## A - Vertex, maximum and minimum values of a quadratic function

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

The term $(x - h)^2$ is a square, hence is either positive or equal to zero.

$(x - h)^2 \ge 0$

If you multiply both sides of the above inequality by coefficient a, there are two possibilities to consider, a is positive or a is negative.

case 1: $a$ is positive
$a(x - h)^2 \ge 0$.

Add $k$ to the left and right sides of the inequality

$a(x - h)^2 + k \ge k$.

The left side represents $f(x)$, hence $f(x) \ge k$. This means that $k$ is the minimum value of function $f$.

case 2: $a$ is negative
$a(x - h)^2 \le 0$.

Add $k$ to the left and right sides of the inequality

$a(x - h)^2 + k \le k$.

The left side represents $f(x)$, hence $f(x) \le k$. This means that $k$ is the maximum value of function $f$.

Note also that $k = f(h)$, hence point $(h,k)$ represents a minimum point when $a$ is positive and a maximum point when $a$ is negative. This point is called the vertex of the graph of $f$.

Example: Find the vertex of the graph of each function and identify it as a minimum or maximum point.
a) $f(x) = -(x + 2)^2 - 1$
b) $f(x) = -x^2 + 2$
c) $f(x) = 2(x - 3)^2$

a) $f(x) = -(x + 2)^2 - 1 = -(x - (-2))^2 - 1$
$a = -1$ , $h = -2$ and $k = -1$. The vertex is at $(-2,-1)$ and it is a maximum point since $a$ is negative.

b) $f(x) = -x^2 + 2 = -(x - 0)^2 + 2$
$a = -1$ , $h = 0$ and $k = 2$. The vertex is at $(0,2)$ and it is a maximum point since $a$ is negative.

c) $f(x) = 2(x - 3)^2 = 2(x - 3))^2 + 0$
$a = 2$ , $h = 3$ and $k = 0$. The vertex is at $(3,0)$ and it is a minimum point since $a$ is positive.

Interactive Tutorial

Use the html 5 (better viewed using chrome, firefox, IE 9 or above) applet below to explore the graph of a quadratic function in vertex form: $f(x)=a (x-h)^2+k$ where the coefficients $a$, $h$ and $k$ may be changed in the applet below. Enter values in the boxes for $a$, $h$ and $k$ and press draw.

 a = 1 -10+10 h = 0 -10+10 k = 1 -10+10
>

1 - Use the boxes on the left panel of the applet to set $a$ to $-1$, $h$ to $-2$ and $k$ to $1$. Check the position of the vertex and whether it is a minimum or a maximum point. Compare to part a) in the example above.

2 - Set $a$ to $-1$, $h$ to $0$ and $k$ to $2$. Check the position of the vertex and whether it is a minimum or a maximum point. Compare to part b) in the example above.

3 - Set $a$ to $2$, $h$ to $3$ and $k$ to $0$. Check the position of the vertex and whether it is a minimum or a maximum point. Compare to part c) in the example above.

4 - Set $h$ and $k$ to some values and $a$ to positive values only. Check that the vertex is always a minimum point.

5 - Set $h$ and $k$ to some values and $a$ to negative values only. Check that the vertex is always a maximum point.

## B - x intercepts of the graph of a quadratic function in standard form

The x intercepts of the graph of a quadratic function $f$ given by
$f(x) = a(x - h)^2 + k$

are the real solutions, if they exist, of the quadratic equation
$a(x - h)^2 + k = 0$

add $-k$ to both sides

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

divide both sides by $a$

$(x - h)^2 = -k/a$

The above equation has real solutions if $-\dfrac{k}{a}$ is positive or zero.

The solutions are given by

$x_1 = h + \sqrt{-\dfrac{k}{a}}$
$x_2 = h - \sqrt{-\dfrac{k}{a}}$

Example: Find the $x$ intercepts for the graph of each function given below

a) $f(x) = -2(x - 3)^2 + 2$
b) $g(x) = -(x + 2)^2$
c) $h(x) = 4(x - 1)^2 + 5$

a) To find the $x$ intercepts, we solve

$-2(x - 3)^2 + 2 = 0$

$-2(x - 3)^2 = -2$

$(x - 3)^2 = 1$

two real solutions $x_1 = 3 + \sqrt 1 = 4$ and $x_2 = 3 - \sqrt 1 = 2$

The graph of function in part a) has two $x$ intercepts are at the points $(4,0)$ and $(2,0)$

b) We solve

$-(x + 2)^2 = 0$

one repeated real solution $x_1 = - 2$

The graph of function in part b) has one $x$ intercept at $(-2,0)$.

c) We solve

$4(x - 1)^2 + 5 = 0$

$-k/a = -5/4$ is negative. The above equation has no real solutions and the graph of function $h$ has no $x$ intercept.

Interactive Tutorial

1 - Go back to the applet window and set the values of $a$, $h$ and $k$ for each of the examples in parts a, b and c above and check the the $x$ intercepts of the corresponding graphs.

2 - Use the applet window to find any x intercepts for the following functions. Use the analytical method described in the above example to find the $x$ intercepts and compare the results.
a) $f(x) = 5(x - 3)^2 + 3$
b) $g(x) = -(x + 2)^2 + 1$
c) $h(x) = 3(x - 1)^2$

3 - Use the applet window and set $a$ and $k$ to values such that $-k/a \lt 0$. How many $x$-intercepts the graph of $f$ has ?

4 - Use the applet window and set $k$ to zero. How many $x$-intercepts the graph of $f$ has ?

5 - Use the applet window and set $a$ and $k$ to values such that $-k/a \gt 0$. How many $x$-intercepts the graph of $f$ has ?

## C - From vertex form to general form with $a$, $b$ and $c$.

Rewriting the vertex form of a quadratic function into the general form is carried out by expanding the square in the vertex form and grouping like terms.

Example: Rewrite $f(x) = -(x - 2)^2 - 4$ into general form with coefficients $a$, $b$ and $c$.

Expand the square in $f(x)$ and group like terms

$f(x) = -(x - 2)^2 - 4 = -(x^2 -4 x + 4) - 4$

$= - x^2 + 4 x - 8$

A tutorial on how to find the equation of a quadratic function given its graph can be found in this site.

More references on quadratic functions and their properties.

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Updated: 2 April 2013