Chapter SH, Problem 9.8.8EP

To determine

The graph of $y=3x^2+6x+3$ is to be drawn.

Explanation of Solution

Given:

  $y=3x^2+6x+3$

Graph:

  

Interpretation:

The given equation is:

  $y=3x^2+6x+3$

Now, the standard quadratic eq. is given by the following expression:

  $y=A{x}^2+Bx+C$

Where,

  $y$ -intercept: $C$

On comparing the given eq. with the standard eq. get that:

  $\begin{array}{l}C=3\\ B=6\\ A=3\end{array}$

The vertex form of quadratic is given by the following expression:

  $y=a{\left(x-h\right)}^2+k$

Where,

Vertex: $h,k$

Put the given eq. in the form of vertex then:

  $\begin{array}{l}y=a{\left(x-h\right)}^2+k\\ y=3\left(x^2+2\left(-1\right)x+{\left(-1\right)}^2\right)+0\\ y=3{\left(x-1\right)}^2+0\end{array}$

Get that: vertex is at $\left(h,k\right)=\left(-1,0\right)$

So, the parabola eq. $y=3x^2+6x+3$ in the vertex form is drawn as above with the help of the following results:

  $\begin{array}{l}a=3\\ h=-1\\ k=0\end{array}$