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 25E

a.

To determine

To Express: The quadratic function in standard form.

Expert Solution

Answer to Problem 25E

the quadratic function is expressed in standard form as  f(x)=3(x1)22

Explanation of Solution

Given: The function is f(x)=(3x26x+1)

Calculation:

The quadratic function f(x)=(3x26x+1) is expressed in standard form as:

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

The parabola opens upwards if a>0 .

Solve the function:

  f(x)=(3x26x+1) f(x)=3(x22x)+1                     [factor 3 the x terms] f(x)=3(x22x+1)+1(31)    [complete the square: add 1 to parentheses, subtract (31) outside] f(x)=3(x1)22                       [factor and multiply]

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

Therefore, the quadratic function is expressed in standard form as  f(x)=3(x1)22

b.

To determine

To Sketch: The graph of the quadratic function.

Expert Solution

Explanation of Solution

Given: The function is f(x)=(3x26x+1)

Graph:

The standard form of the function is:

   f(x)=3(x1)22

From the standard form it is observed that the graph is a parabola that opens upward and has vertex (1,2) . As an aid to sketching the graph, find the intercepts.

The y-intercept=f(0)=1 and The x-intercept is 1±63 and 0 . The graph f is sketched in the figure below.

Use graphing calculator to graph the function: f(x)=(3x26x+1)

  Precalculus: Mathematics for Calculus - 6th Edition, Chapter 3.1, Problem 25E

c.

To determine

To Find: The maximum or minimum value of the function.

Expert Solution

Answer to Problem 25E

The value of minima is f(1)=(2)

Explanation of Solution

Given: The function is f(x)=(3x26x+1)

Calculation:

From the above graph it is seen that the parabola opens upward, since the coefficient of x2 is positive, f has minimum value. The value of minima is f(1)=(2)

Have a homework question?

Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!