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 6E

a.

To determine

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

Expert Solution

Answer to Problem 6E

The coordinates of the vertex are (2,8)

Explanation of Solution

Given: The function is f(x)=12x2+2x+6

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)=12x2+2x+6f(x)=12(x22x)+6                           [factor 12 from the x terms] f(x)=12(x22x+4)+6(12)4    [complete the square : add +4 inside parentheses, subtract (12)4 outside]  f( x )= 1 2 (x2) 2 +8                             [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)=(2,8)

b.

To determine

To Find: The maximum or minimum value of f.

Expert Solution

Answer to Problem 6E

The maximum value is f(2)=8

Explanation of Solution

Given:The function is f(x)=12x2+2x+6

Calculation:The standard form of the function is:

  f(x)=12(x2)2+8

If a<0 , then the maximum value of f is f(h)=k , and is represented as f(2)=8

Since, the coefficient of x2 is negative, f has a maximum value. The maximum value is f(2)=8

c.

To determine

To Find: The domain and range of f.

Expert Solution

Answer to Problem 6E

the domain is from (,) and the range is (,8]

Explanation of Solution

Given: The function is f(x)=12x2+2x+6

Calculation:

The standard form of the function is:

  f(x)=12(x2)2+8

It is observed from the standard form of the function thatthe domain is from (,) and the range is (,8]

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