Chapter 7.6, Problem 35WE

a.

To determine

The vertex of the graph of the function $g\left(x\right)=2-\frac{1}{2}{\left(x+1\right)}^2$ .

Answer to Problem 35WE

The vertex of graph of function is $\left(-1,2\right)$ .

Explanation of Solution

Given information: The given function is $g\left(x\right)=2-\frac{1}{2}{\left(x+1\right)}^2$ .

Calculation: The given function can be written as,

  $\begin{array}{c}y=2-\frac{1}{2}{\left(x+1\right)}^2\mathrm{....}\left(\text{Replace }g\left(x\right)=y\right)\\ \text{or, }y-2=-\frac{1}{2}{\left(x+1\right)}^2\mathrm{....}\left(1\right)\end{array}$

Since (1) represents a parabola of the form $y-k=a{\left(x-h\right)}^2$ . Therefore the graph for the function has vertex $\left(-1,2\right)$ .

b.

To determine

The domain of the function $g\left(x\right)=2-\frac{1}{2}{\left(x+1\right)}^2$ .

Answer to Problem 35WE

The domain of the function is all real numbers.

Explanation of Solution

Given information: The given function is $g\left(x\right)=2-\frac{1}{2}{\left(x+1\right)}^2$ .

Calculation: From (1), the parabola obtained represents a parabola opening downwards because of negative coefficient of ${\left(x+1\right)}^2$ . Therefore the domain of the function has no restrictions at all.

Hence the domain of the function is all real numbers.

c.

To determine

The range of the function $g\left(x\right)=2-\frac{1}{2}{\left(x+1\right)}^2$ .

Answer to Problem 35WE

The range of the function is $\left\{y|y\le 2\right\}$ .

Explanation of Solution

Given information: The given function is $g\left(x\right)=2-\frac{1}{2}{\left(x+1\right)}^2$ .

Calculation: From (1), the parabola obtained represents a parabola opening downwards because of negative coefficient of ${\left(x+1\right)}^2$ . Also the parabola has the vertex $\left(-1,2\right)$ which means the upper limit for range is $2$ . Since, the parabola (1) is upper bound. So its maximum range will be $\left(y\right)$ coordinate of its vertex.

Hence the range of the function is all real numbers less than or equal to $2$ . Mathematically, the range of function is $\left\{y|y\le 2\right\}$ .

d.

To determine

The zeros of the function $g\left(x\right)=2-\frac{1}{2}{\left(x+1\right)}^2$ .

Answer to Problem 35WE

The zeros of the function are $1,-3$ .

Explanation of Solution

Given information: The given function is $g\left(x\right)=2-\frac{1}{2}{\left(x+1\right)}^2$ .

Calculation: The zeros of the function can be obtained by equating the given function equal to zero. Therefore $g\left(x\right)=0$ gives,

  $\begin{array}{c}2-\frac{1}{2}{\left(x+1\right)}^2=0\\ \text{or, }{\left(x+1\right)}^2=4\\ \left(x+1\right)=\pm \sqrt{4}\\ x+1=\pm 2\\ x=1,-3\end{array}$

Hence the zeros of the function are $1,-3$ .