1.4 More Quadratic Functions and Applications 1 In the previous section, we transformed a quadratic function from the form f(x)=ax 2 +bx+c to the form

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1.4 More Quadratic Functions and Applications

In the previous section, we transformed a quadratic function from the form f(x)=ax2+bx+c to the form f(x)=a(x–h)2+k in order to more easily identify the vertex, (h,k). We did this by the process known as completing the square. In this section we will show you another way to perform this transformation.

Using the process of completing the square, we can obtain the vertex with the following procedure:

1. Identify the real number values a, b, and c.If a is positive, the parabola will open upward.If a is negative, the parabola will open downward.

b2. Find . This is the x coordinate of the vertex.

2a

b3. Find f . This is the y coordinate of the vertex.

2a

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1. Identify a, b, and c.

b2. Find x= .

2a

b3. Find f .

2a

2 Find the vertex of Example 1. f(x) = 2x 8x 5 using x =-b/2a.

Solution: a=2, b=-8, and c=5

( 8)x 2

2(2)

2f(2) = 2(2) 8(2) 5 3

Answer: The vertex is (2,-3).

Your Turn Problem #1

2Find the vertex of f(x) = 3x 18x 22 using x =-b/2a. Answer: (-3,-5)

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Example 2. Graph f(x)= –3x2+6x+2.

Solution:

b (6)x 1

2a 2( 3)

The vertex is (1,5).

1. Find the vertex.

2f(1) 3(1) 6(1) 2 5

2. Make a table. Use two numbers to the right of 1 and two numbers to the left

of 1.

x y 2 2 3 -7 0 2-1 -7

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2Graph f(x) = 2x 12x 14 using x =-b/2a.

Vertex (3,-4)

x y 4 -2 5 4 2 -2 1 4

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A parabola can have two x intercepts (figure 1), one x intercept (figure 2), or no x intercepts (figure 3). For those that do, we usually want to know what they are.

To find x intercepts: set the function equal to 0, and solve the resulting equation for x.

Note: If the values of x found by setting the equation equal to 0 are not real, then the parabola has no x intercepts.

x Intercepts of a Parabola

x

f(x)

x

f(x)

Figure 1 Figure 2 Figure 3

x

f(x)

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Example 3.Find the x intercepts and the vertex of f(x)= –x2+6x–5.

Solution: To find the x intercepts, set the function equal to zero and solve.

The x-intercepts are (5,0) and (1,0).

–x2 + 6x – 5 = 0

(x – 5)(x – 1) = 0

x – 5=0

x=5

x – 1=0

x=1

x2 –6x +5 = 0To find the vertex, find x = -b/2a.

a= –1, b=6, and c= –5

(6)x 3

2( 1)

2f(3) = (3) 6(3) 5

f(3) = 9 18 5 4

The vertex is 3,4 .

Find the x-intercepts and the vertex of f(x)=2x2 – 36x+160.


Answer: The x-intercepts are (8,0) and (10,0).

vertex 9, 2

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There is a correspondence between x-intercepts and zeros, which are x-values that have corresponding y-values that equal 0. 2 and –3 are zeros of the function f(x)=x2+x–6.The graph of this quadratic function is a parabola with x-intercepts of (2,0) and (-3,0).

The procedure for finding zeros of a function is the same as that of finding x-intercepts for the graph of a function, except the answer is not an ordered pair, but instead, merely the x-value.Example 4.Find the zeros of f(x)= –x2+6x+6.

Solution: To find the zeros, set the function equal to zero and solve. This trinomial does not factor. Use quadratic formula or completing the square to solve.

–x2+6x–6=0

x2 – 6x– 6 = 0

2( 6) ( 6) 4(1)( 6)x

2(1)

6 60x

2

6 2 15x 3 15

2

The zeros of the function are:3 15 and 3 15.

Find the zeros of f(x)=12x2 – x – 20.


4 5Answer: The zeros are and .

3 4

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Applications of Quadratic Functions

When quadratic functions are used to model real life situations, we see that themodel will either have a lowest or highest value at the vertex. 1. The x-value of the vertex indicates where the minimum or maximum occurs.

2. F(x) gives the minimum or maximum value of the function.

x

f(x)

x

f(x)

minimum

maximum

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Example 5. Dog Homes R Us earns a yearly profit according to the functionf(x)= –0.4x2+80x–200, where x is the number of dog houses produced and

sold and f(x) is the profit in thousands of dollars. Find the maximum profit the company can earn in a week. Solution: Since this is a quadratic function, in which the value of ‘a’ is negative, we know that the parabola of the function’s graph opens downward. We are therefore guaranteed a maximum value, which would correspond to the function’s maximum profit. Understand that the x-value of the vertex would give the number of dog houses produced and sold that would yield the maximum profit. It is the y-value of the vertex that is the maximum profit. (Recall this value is in thousands of dollars.)

a= –0.4, b=80, c= –200

(80)x 100

2( 0.4)

2f(100) = 0.4(100) 80(100) 20

f(100) =3800

Remember, this maximum value is in thousands of dollars, so 3800 thousand dollars is more commonly known as $3,800,000.

Layton Ltd. earns a weekly profit according to the function f(x)= –0.001x2+2.45x–525, where x is the number of salt shakers produced and sold and f(x) is the profit in thousands of dollars. Find the maximum profit the corporation can earn in a year.


Answer: The maximum profit is $975,625.

The EndB.R.1-5-07

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1.4 More Quadratic Functions and Applications 1 In the previous section, we transformed a quadratic function from the form f(x)=ax 2 +bx+c to the form