Chapter 4.7, Problem 7CYU

To determine

To graph: the quadratic equation as mentioned in the problem.

Explanation of Solution

Given information:

The quadratic equation in vertex form is $y=\frac{1}{2}{\left(x+6\right)}^2-8$ .

Graph:

The graph mentioned below is in real number line of x axis and y axis.

Where vertex $V\left(-6,-8\right)$ .

  

Interpretation:

Consider the quadratic equation in vertex form $y=\frac{1}{2}{\left(x+6\right)}^2-8$ .

The mentioned quadratic equation above is in vertex form $y=a{\left(x-h\right)}^2+k$ where $\left(h,k\right)$ is the vertex.

Therefore the vertex of equation is $V\left(-6,-8\right)$ .

We can write the mentioned quadratic equation in the standard form as $\begin{array}{l}y=\frac{1}{2}{\left(x+6\right)}^2-8\Rightarrow y=\frac{1}{2}\left(x^2+36+12x\right)-8\Rightarrow y=\frac{1}{2}{x}^2+18+6x-8\\ y=\frac{1}{2}{x}^2+6x+10.\end{array}$

Which is comparable to $a{x}^2+bx+c$ Where $a=\frac{1}{2}$ which is greater than $0$ .

Since $a>0$ the direction of graph will be positive side of y axis.

The discriminant of standard form of quadratic equation $y=\frac{1}{2}{x}^2+6x+10$ is $D=b^2-4*a*c\Rightarrow d={\left(6\right)}^2-4*\left(\frac{1}{2}\right)*\left(10\right)=36-20=16$ .

Here $D>0$ .

Therefore the graph will lie above x axis.

At $x=0$ the $y=\frac{1}{2}{x}^2+6x+10$ will yield $y=10$ .

Therefore graph will cut y axis at $\left(0,10\right)$ .

Therefore the graph will be as mentioned below.