BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

Solutions

Chapter 3.1, Problem 5E

a.

To determine

To Find: The coordinates of the vertex of a quadratic function f.

Expert Solution

Answer to Problem 5E

The coordinates of the vertex are (3,4)

Explanation of Solution

Given: The function is f(x)=x2+6x5

Calculation:

The quadratic function f(x)=x2+6x5 is expressed in standard form as:

  f(x)=a(xh)2+k , by completing the square. The graph of f is a parabola with vertex (h,k) ,

The parabola opens downward if a<0 .

Solve the function:

  f(x)=x2+6x5        =1(x26x)5                      [factor 1 from the x terms]        =1(x26x+9)5(1)9    [complete the square : add +9 inside parentheses, subtract (1)9 outside]         =1 (x3) 2 +4                        [factor and multiply]

On comparing the above equation with standard form f(x)=a(xh)2+k ,

Therefore, the coordinates of the vertex are (h,k)=(3,4)

b.

To determine

To Find: The maximum or minimum value of f.

Expert Solution

Answer to Problem 5E

the coefficient of x2 is negative, f has a maximum value. The maximum value is f(3)=4

Explanation of Solution

Given:The function is f(x)=x2+6x5

Calculation:The standard form of the function is:

   f(x)=1(x3)2+4

If a<0 , then the maximum value of f is f(h)=k , and is represented as f(3)=4

Since, the coefficient of x2 is negative, f has a maximum value. The maximum value is f(3)=4

c.

To determine

To Find: The domain and range of f.

Expert Solution

Answer to Problem 5E

the domain is from (,) and the range is (,4]

Explanation of Solution

Given: The function is f(x)=x2+6x5

Calculation:

The standard form of the function is:

   f(x)=1(x3)2+4

It is observed from the standard form of the function thatthe domain is from (,) and the range is (,4]

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