Chapter 12.6, Problem 4E

To determine

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

Explanation of Solution

Given information:

The quadratic equation is $y=x^2+2x-15$ .

Graph:

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

Where vertex $V\left(-1,-16\right)$ .

  

Interpretation:

Consider the quadratic equation $y=x^2+2x-15$ .

The mentioned quadratic equation above is in standard form $a{x}^2+bx+c$ where $a=1,b=2,c=-15$ .

We can write the mentioned quadratic equation in the vertex form as $y=x^2+2x-15\Rightarrow y=x^2+2*x*1+1-1-15\Rightarrow y={\left(x+1\right)}^2-16$ 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 greater than $0$ .

Therefore the vertex of equation is $V\left(-1,-16\right)$ .

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

The discriminant of standard form of quadratic equation $y=x^2+2x-15$ is $D=b^2-4*a*c\Rightarrow D={\left(2\right)}^2-4*\left(1\right)*\left(-15\right)=64.$ .

Here also $D=64>0$ .

Therefore the graph will lie above the x axis.

At $x=0$ the $y=x^2+2x-15$ will yield $y=-15$ .

Therefore graph will cut y axis at $\left(0,-15\right)$ .

Therefore the graph will be as mentioned below.