Chapter 12.6, Problem 3E

To determine

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

Explanation of Solution

Given information:

The quadratic equation is $y=-x^2+16$ .

Graph:

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

Where vertex $V\left(0,16\right)$ .

  

Interpretation:

Consider the quadratic equation $y=-x^2+16$ .

The mentioned quadratic equation above is in standard form $a{x}^2+bx+c$ where $a=-1,b=0,c=16$ .

We can write the mentioned quadratic equation in the vertex form as $y=-x^2+16\Rightarrow y=-{\left(x-0\right)}^2+16$ 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 less than $0$ .

Therefore the vertex of equation is $V\left(0,16\right)$ .

Since $a<0$ the direction of graph will be negative side of y axis.

The discriminant of standard form of quadratic equation $y=-x^2+16$ is $D=b^2-4*a*c\Rightarrow D={\left(-1\right)}^2-4*\left(-1\right)*\left(0\right)=1.$

Here also $D=1>0$ .

Therefore the graph will lie above the x axis.

At $x=0$ the $y=-x^2+16$ will yield $y=1$ .

Therefore graph will cut y axis at $\left(0,1\right)$ .

Therefore the graph will be as mentioned below.