Chapter 12.6, Problem 2E

To determine

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

Explanation of Solution

Given information:

The quadratic equation is $y=3x+x^2$ .

Graph:

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

Where vertex $V\left(-\frac{3}{2},-\frac{9}{4}\right)$ .

  

Interpretation:

Consider the quadratic equation $y=3x+x^2$ .

The mentioned quadratic equation above is in standard form $a{x}^2+bx+c$ where $a=1,b=3,c=0$ .

We can write the mentioned quadratic equation in the vertex form as $y=3x+x^2\Rightarrow y=x^2+2*x*\frac{3}{2}+{\frac{3}{2}}^2-{\frac{3}{2}}^2\Rightarrow y={\left(x+\frac{3}{2}\right)}^2-\frac{9}{4}$ 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 $V\left(-\frac{3}{2},-\frac{9}{4}\right)$ .

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

The discriminant of standard form of quadratic equation $y=3x+x^2$ is $D=b^2-4*a*c\Rightarrow d={\left(3\right)}^2-4*\left(1\right)*\left(0\right)=9.$ .

Here also $D=9>0$ .

Therefore the graph will lie above the x axis.

At $x=0$ the $y=3x+x^2$ will yield $y=0$ .

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

Therefore the graph will be as mentioned below.