Quadratic Functions in Standard Form

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 ≥ 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 ≥ 0.

Add k to the left and right sides of the inequality

a(x - h)2 + k ≥ k.

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

case 2: a is negative
a(x - h)2 ≤ 0.

Add k to the left and right sides of the inequality

a(x - h)2 + k ≤ k.

The left side represents f(x), hence f(x) ≤ 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) = -x2 + 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) = -x2 + 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 =
-10+10

h =
-10+10

k =
-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 - k / a is positive or zero.

The solutions are given by

x1 = h + √(- k / a)
x2 = h - √(- 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 + √1 = 4 and x2 = 3 - √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 x1 = - 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 < 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 > 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 = -(x2 -4 x + 4) - 4

= - x2 + 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.