Quadratic Functions Help

Introduction to Quadratic Functions

The graph of every quadratic function, f ( x ) = ax ² + bx + c , is a transformation of the graph of y = x ² . (See Figure 6.1)

Quadratic Functions

Fig. 6.1

When the function is written in the form f ( x ) = a ( x − h ) ² + k , we have a pretty good idea of what its graph looks like: h will cause the graph to shift horizontally, and k will cause it to shift vertically. The point (0, 0) on y = x ² has shifted to ( h, k ). This point is the vertex . On a parabola that opens up (when a is positive), the vertex is the lowest point on the graph. The vertex is the highest point on a parabola that opens down (when a is negative).

We need to know the vertex when sketching a parabola. Once we have the vertex, we will find two points to its left and two points to its right. We should choose points in such a way that shows the curvature around the vertex and how fast the ends are going up or down. It does not matter which points we choose, but a good rule of thumb is to find h − 2 a, h − a, h + a , and h + 2 a . Because a parabola is symmetric about the line x = h (the vertical line that goes through the vertex), the y -values for h − a and h + a will be the same and the y -values for h − 2 a and h + 2 a will be the same, too. We will also find the intercepts.

Examples

Sketch the graph for the following quadratic functions. Find the y -intercept and the x -intercepts, if any.

f ( x ) = ( x − 1) ² − 4

a = 1, h = 1, k = −4 The parabola opens up and the vertex is (1, −4). For the y -intercept, let x = 0 in the function. The y -intercept is (0−1) ² −4 = 3. For the x -intercepts, let y = 0 and solve for x .

( x − 1) ² − 4 = 0

( x − 1) ² = 4 Take the square root of each side.

x − 1 = ±2

x = 1 ± 2 = 1 + 2, 1−2 = 3, −1

The x -intercepts are 3 and − 1.

Table 6.1

	x	y	Plot this point
h − 2 a	1 − 2(1) = −1	(−1 − 1) ² − 4 = 0	(−1, 0)
h − a	1 − 1 = 0	(0 − 1) ² − 4 = −3	(0,−3)
h	1	−4	(1,−4)
h + a	1 + 1 = 2	(2 − 1) ² − 4 = −3	(2,−3)
h + 2 a	1 + 2(1) = 3	(3 − 1) ² − 4 = 0	(3, 0)

Quadratic Functions Examples

Fig. 6.2

g ( x ) = −2( x + 1) ² + 18

a = −2, h = −1, k = 18 The parabola opens down, and the vertex is (−1, 18).

y = −2(0 + 1) ² + 18 y = 16	−2( x + 1) ² + 18 = 0
y = 16	− 2( x + 1) ² = −18
	( x + 1) ² = 9
	x + 1 = ±3
	x = −1 ±3 = −1 − 3,
	−1 + 3 = −4, 2

The y -intercept is 16 and the x -intercepts are −4 and 2.

Table 6.2

	x	y	Plot this point
h − 2 a	−1 − 2(−2) = 3	−2(3 + 1) ² + 18 = −14	(3,−14)
h − a	−1 − (−2) = 1	−2(1 + 1) ² + 18 = 10	(1, 10)
h	−1	18	(−1, 18)
h + a	−1 + (−2) = −3	−2(−3 + 1) ² + 18 = 10	(−3, 10)
h + 2 a	−1 + 2(−2) = −5	−2(−5 + 1) ² + 18 = −14	(−5,−14)

Quadratic Functions Examples

Fig. 6.3

Quadratic Functions Examples

Quadratic Functions Examples The parabola opens up, and the vertex is (−1, 2). Because the parabola opens up ( is positive) and the vertex is above the x -axis ( k = 2 is positive), there will be no x -intercept. If we were to solve the equation , we would not get a real number solution. The y -intercept is Quadratic Functions Examples .

Table 6.3

	x	y	*Plot this point*
h − 2 a	−1 − 2( ) = −2	(−2 + 1) ² + 2 = 2	(−2, 2 )
h − a	−1 − = −1	(−1 + 1) ² + 2 = 2	(−1 , 2 )
h	−1	2	(−1, 2)
h + a	−1 + = −	(− + 1) ² + 2 = 2	(− , 2 )
h + 2 a	−1 + 2( ) = 0	(0 + 1) ² + 2 = 2	(0, 2 )

Quadratic Functions Examples

Fig. 6.4

Finding the Equation for Quadratic Functions

By knowing the vertex and one other point on the graph, we can find an equation for the quadratic function. Once we know the vertex, we have h and k in y = a ( x − h ) ² + k . By using another point (x, y), we can find a.

Example

The vertex for a quadratic function is (−3, 4), and the y -intercept is −10. Find an equation for this function.

Let h = −3, k = 4 in y = a ( x − h ) ² + k to get y = a ( x + 3) ² + 4. Saying that the y -intercept is −10 is another way of saying (0, −10) is a point on the graph. We can let x = 0 and y = −10 in y = a ( x + 3) ² + 4 to find a.

One equation for this function is Quadratic Functions Example .

Completing the Square

Quadratic equations are not normally written in the convenient form f ( x ) = a ( x − h ) ² + k . We can complete the square on f ( x ) = ax ² + bx + c to find ( h, k ). Begin by completing the square on the x ² and x terms.

Examples

Find the vertex.

y = x ² − 6 x − 2

We need to balance putting +(6/2) ² = 9 in the parentheses by adding −9 to −2

y = ( x ² − 6 x + 9) − 2 − 9

y = ( x − 3) ² − 11 The vertex is (3, −11).
f ( x ) = 4 x ² + 8 x + 1

We will begin by factoring a = 4 from 4 x ² + 8 x. Then we will complete the square on the x ² and x terms.

f ( x ) = 4 x ² + 8 x + 1

f ( x ) = 4( x ² + 2 x ) + 1

f ( x ) = 4( x ² + 2 x + 1) + 1+?

By putting +1 in the parentheses, we are adding 4(1) = 4. We need to balance this by adding −4 to 1.

f ( x ) = 4( x ² + 2 x + 1) + 1 + (−4)

f ( x ) = 4( x + 1) ² − 3 The vertex is (−1, −3).

Finding the Terms in Parentheses

When factoring an unusual quantity from two or more terms, it is not obvious what terms go in the parentheses. We can find the terms that go in the parentheses by writing the terms to be factored as numerators of fractions and the number to be factored as the denominator. The terms that go in the parentheses are the simplified fractions.

We need to factor a = −3 from − 3 x ² + 9 x.

By putting in the parentheses, we are adding . We need to balance this by adding
Factoring , we have

By adding in the parentheses, we are adding . We need to balance this by adding −3/8 to −2.

Shortcuts to Finding the Equation for Quadratic Functions

One advantage to the form f ( x ) = ax ² + bx + c is that it is usually easier to use to find the intercepts. We can use factoring and the quadratic formula when it is in this form. Also, c is the y -intercept. Because a is the same number in both forms, we can tell whether the parabola opens up or down when the equation is in either form. It can be tedious to complete the square on f ( x ) = ax ² + bx + c to find the vertex. Fortunately, there is a shortcut.

Quadratic Functions Examples

This shortcut comes from completing the square to rewrite f ( x ) = ax ² + bx + c as f ( x ) = a ( x − h ) ² + k .

Quadratic Functions Examples

It is easier to find k by evaluating the function at Quadratic Functions Examples than by using this formula.

Example

Use the shortcut to find the vertex.

f ( x ) = −3 x ² + 9 x + 4

Finding the Maximum/Minimum Value for a Function

An important topic in calculus is optimizing functions; that is, finding a maximum and/or minimum value for the function. Precalculus students can use algebra to optimize quadratic functions. A quadratic function has a minimum value (if its graph opens up) or a maximum value (if its graph opens down). If a is positive, then k is the minimum functional value. If a is negative, then k is the maximum functional value. These values occur at x = h.

Examples

Find the minimum or maximum functional value and where it occurs.

f ( x ) = −( x − 3) ² + 25

The parabola opens down because a = −1 is negative. This means that k = 25 is the maximum functional value. It occurs at x = 3.
y = 0.01 x ² − 6 x + 2000

a = 0.01 is positive, so k = 1100 is the minimum functional value. The minimum occurs at x = 300.