Chapter SH, Problem 9.8.7EP

To determine

The graph of $y=5x^2-20x+37$ is to be drawn.

Explanation of Solution

Given:

  $y=5x^2-20x+37$

Graph:

  

Interpretation:

The given equation is:

  $y=5x^2-20x+37$

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:

  $\begin{array}{l}C=37\\ B=-20\\ A=5\end{array}$

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:

  $\begin{array}{l}y=a{\left(x-h\right)}^2+k\\ y=5x^2-20x+20+17\\ y=5\left(x^2-4x+4\right)+17\\ y=5{\left(x-2\right)}^2+17\end{array}$

Get that: vertex is at $\left(h,k\right)=\left(2,17\right)$

So, the parabola eq. $y=5x^2-20x+37$ in the vertex form is drawn as above with the help of the following results:

  $\begin{array}{l}a=5\\ h=2\\ k=17\end{array}$