Chapter 8, Problem 8CA

a.

To determine

To complete: The function $f(x)=5{(x\_3)}^2\_4$ where vertex is (-3, 4) using the symbols

Answer to Problem 8CA

  $f(x)=5{(x+3)}^2+4$

Explanation of Solution

Given information: Vertex of the function is (-3, 4)

Formula used:

The vertex form of the quadratic equation is given by

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

Use the given vertex and complete the function using symbols

  $\begin{array}{l}$ f(x)=a{(x-h)}^2+k$(h,k):(-3,4)\\ $ f(x)=5{(x-(-3))}^2+4$$ f(x)=5{(x+3)}^2+4$\end{array}$

b.

To determine

To complete: The function $g(x)=-(x\_2)(x\_8)$ where x − intercept is − 8 and 2.

Answer to Problem 8CA

  $g(x)=-(x-2)(x+8)$

Explanation of Solution

Given information: x intercepts of g(x) are -8 and 2

Formula used:

Use the factor form of the quadratic equation

  $\begin{array}{l}g(x)=(x-a)(x-b)\\ a,b:\text{intercepts}\end{array}$

Calculation:

The given intercept are -8 and 2, that is the value of a and b

  $\begin{array}{l}g(x)=-(x-a)(x-b)\\ a=2,b=-8\\ g(x)=-(x-2)(x+8)\end{array}$

c.

To determine

To complete: The function $h(x)=\_3x^2\_6$ where range is $y\ge -6$ .

Answer to Problem 8CA

  $h(x)=+3x^2-6$

Explanation of Solution

Given information: The range of the function is greater than or equal to -6

Calculation:

When the range of the function is greater than -6, then the graph of the function goes up.

So the $x^2$ term is positive

Range is greater than -6 .The value of 6 is negative

  $h(x)=+3x^2-6$

d.

To determine

To complete: The function $j(x)=\_4(x\_1)(x\_1)$ where range is $y\le 4$

Answer to Problem 8CA

  $j(x)=-4(x-1)(x+1)$

Explanation of Solution

Given information: The range of the function is less than or equal to 4

Calculation:

When the range of the function is less than 4, then the graph of the function goes down

So the first term is negative

Lets fill in the other two boxes by trial and error method. Plug in negative for both

  $\begin{array}{l}j(x)=-4(x-1)(x-1)\\ $ j(x)=-4{(x-1)}^2$\text{Vertex:}(1,0)\\ \text{range}:y\le 0\end{array}$

The range becomes less than 0. So we change the sign and check

  $\begin{array}{l}j(x)=-4(x-1)(x+1)\\ j(x)=-4(x^2-1)\\ j(x)=-4x^2+4\\ \text{Vertex:}(0,4)\\ \text{range}:y\le 4\end{array}$

The function is

  $j(x)=-4(x-1)(x+1)$