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 27E

a.

To determine

To Express: The quadratic function in standard form.

Expert Solution

Answer to Problem 27E

the quadratic function is expressed in standard form as f(x)=(x+32)2+214

Explanation of Solution

Given: The function is f(x)=(x23x+3)

Calculation:

The quadratic function f(x)=(x23x+3) 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 downwards if a<0 .

Solve the function:

  f(x)=(x23x+3)f(x)=14(4x212x+12)                  [Multiply and divide by 4]f(x)=14(4x212x9+9+12)       [add and subtract 9]f(x)=14[(4x2+12x+9)21]f(x)=14[(2x+3)221]f(x)=(2x+3)24+214f(x)=(2x+32)2+214f(x)=(x+32)2+214 

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)=(x+32)2+214

b.

To determine

To Sketch: The graph of the quadratic function.

Expert Solution

Explanation of Solution

Given: The function is f(x)=(x23x+3)

Graph:

The standard form of the function is:

  f(x)=(x+32)2+214

From the standard form it is observed that the graph is a parabola that opens upward and has vertex (32,214) . As an aid to sketching the graph, find the intercepts.

The y-intercept=f(0)=3 and The x-intercept is 32±212 and 0 . The graph f is sketched in the figure below.

Use graphing calculator to graph the function: f(x)=(x23x+3)

  Precalculus: Mathematics for Calculus - 6th Edition, Chapter 3.1, Problem 27E

c.

To determine

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

Expert Solution

Answer to Problem 27E

The value of minima is f(32)=(214)

Explanation of Solution

Given: The function is f(x)=(x23x+3)

Calculation:

From the above graph it is seen that the parabola opens downward, since the coefficient of x2 is positive, f has minimum value. The value of minima is f(32)=(214)

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