Chapter 5.7, Problem 30PPS

To determine

To graph: the function as mentioned in the problem.

Explanation of Solution

Given information:

The function is $y=\frac{1}{6}{\left(x-3\right)}^2-10$ .

Graph:

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

Where vertex $V\left(3,-10\right)$ .

  

Interpretation:

Consider the function as mentioned $y=\frac{1}{6}{\left(x-3\right)}^2-10$ .

The mentioned function 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 $V\left(3,-10\right)$ .

We can write the mentioned quadratic equation in the standard form as $y=\frac{1}{6}{\left(x-3\right)}^2-10\Rightarrow y=\frac{1}{6}\left(x^2+9-2*x*3\right)-10\Rightarrow y=\frac{1}{6}{x}^2-x+\frac{3}{2}-10\Rightarrow y=\frac{1}{6}{x}^2-x-\frac{17}{2}$ which is comparable to $a{x}^2+bx+c$ where $a=\frac{1}{6}$ 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 $y=\frac{1}{6}{x}^2-x-\frac{17}{2}$ is $D=b^2-4*a*c\Rightarrow d={\left(-1\right)}^2+4*\left(\frac{1}{6}\right)*\left(\frac{17}{2}\right)=1+\frac{17}{3}=\frac{20}{3}$ .

Here also $D=\frac{20}{3}>0$ .

Therefore the graph will lie above the x axis.

At $x=0$ the $y=\frac{1}{6}{x}^2-x-\frac{17}{2}$ will yield $y=-\frac{17}{2}$ .

Therefore graph will cut y axis at $\left(0,-\frac{17}{2}\right)$ .

Therefore the graph will be as mentioned below.