Chapter 8, Problem 4CT

(a)

To determine

To find: Use the graph to get the domain, range and zeros

Answer to Problem 4CT

  $\begin{array}{l}Domain:-\infty <x<\infty \\ Range:f(x)\le 8\\ x-\mathrm{int}ercepts:3,7\end{array}$

Explanation of Solution

Given information: The graph of the function with some coordinates

Concept used:

The domain is the set of x values for which the function is defined and the value is real.

Range is the set of all y values for which the function is defined.

Calculation:

In the graph there is no restriction for x . The curve is for all value of x. So, domain is set of all real numbers

  $Domain:-\infty <x<\infty$

The graph starts at 8 on y and it goes down

  $Range:f(x)\le 8$

x intercepts are the values where the graph crosses x- axis

  $x-\mathrm{int}ercepts:3,7$

(b)

To determine

To find: Use the graph to get the standard form of the function

Answer to Problem 4CT

  $f(x)=-2x^2+20x-42$

Explanation of Solution

Given information: The graph of the function with some coordinates

  $\begin{array}{l}x-\text{intercepts}:(3,0),(7,0)\\ \text{Vertex}(5,8)\end{array}$

Formula used:

  $\begin{array}{l}$ f(x)=a{(x-h)}^2+k$vertex(h,k)\\ f(x)=a{x}^2+bx+c\end{array}$

Calculation:

Substitute the given vertex in general form

  $\begin{array}{}$ f(x)=a{(x-h)}^2+k$$ \text{vertex}\text{=}\text{(}5,8)$$ f(x)=a{(x-5)}^2+8$\end{array}$

To find the value of ‘ a’ use the ‘x’ intercept

  $\begin{array}{l}$ f(x)=a{(x-5)}^2+8$x-\text{intercept}:(3,0)\\ $ 0=a{(3-5)}^2+8$0=4a+8\\ -8=4a\\ a=-2\end{array}$

Putting value of a back in the equation we get,

  $\begin{array}{l}$ f(x)=-2{(x-5)}^2+8$f(x)=-2(x^2-10x+25)+8\\ f(x)=-2x^2+20x-42\end{array}$

Conclusion: The equation of the graph is $f(x)=-2x^2+20x-42$ .

(c )

To determine

To compare: The graph of f(x) with the function $g(x)=x^2$

Answer to Problem 4CT

f(x) is upside down, f(x) is vertically stretched by 2 units, f(x) shifted up by 8 units and right by 5 units

Explanation of Solution

Given information: The equation obtained from part (b)

  $f(x)=-2{(x-5)}^2+8$

Calculation:

Compare both equations f(x) and $g(x)=x^2$

  $\begin{array}{l}$ f(x)=-2{(x-5)}^2+8$g(x)=x^2\end{array}$

For -2 , the graph is stretched vertically and parabola is upside down

5 is subtracted from x , the graph of f(x) is shifted 5 units right

8 is subtracted from x , the graph of f(x) is shifted 8 units up

(d)

To determine

To graph: Graph the function$h(x)=f(x-6)$

Answer to Problem 4CT

  

Explanation of Solution

Given information:

  $\begin{array}{l}h(x)=f(x-6)\\ $ f(x)=-2{(x-5)}^2+8$\end{array}$

Calculation:

6 is subtracted from x in f(x). So the graph of h(x) will be shifted 6 units to the right

Move all the points on the graph of f(x) , 6 units to the right