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# Quadratic Functions Help (page 2)

based on 4 ratings
By McGraw-Hill Professional
Updated on Oct 4, 2011

## 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 ( xh ) 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 ( xh ) 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 .

### Completing the Square

Quadratic equations are not normally written in the convenient form f ( x ) = a ( xh ) 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.

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

It is easier to find k by evaluating the function at 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.

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