Chapter 12.6, Problem 5E

To determine

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

Explanation of Solution

Given information:

The quadratic equation is $y=x^2-8x+12$ .

Graph:

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

Where vertex $V\left(4,-4\right)$ .

  

Interpretation:

Consider the quadratic equation $y=x^2-8x+12$ .

The mentioned quadratic equation above is in standard form $a{x}^2+bx+c$ where $a=1,b=-8,c=12$ .

We can write the mentioned quadratic equation in the vertex form as $y=x^2-8x+12\Rightarrow y=x^2-2*x*4+16-16+12\Rightarrow y={\left(x-4\right)}^2-4$ which is comparable to $y=a{\left(x-h\right)}^2+k$ where $\left(h,k\right)$ is the vertex and $a=1$ which is greater than $0$ .

Therefore the vertex of equation is $V\left(4,-4\right)$ .

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

The discriminant of standard form of quadratic equation $y=x^2-8x+12$ is $D=b^2-4*a*c\Rightarrow D={\left(-8\right)}^2-4*\left(1\right)*\left(12\right)=64-48=16.$

Here also $D=16>0$ .

Therefore the graph will lie above the x axis.

At $x=0$ the $y=x^2-8x+12$ will yield $y=12$ .

Therefore graph will cut y axis at $\left(0,12\right)$ .

Therefore the graph will be as mentioned below.