Chapter SH, Problem 9.1.7EP

To determine

The vertex, equation of axis of symmetry and y-intercept.

Answer to Problem 9.1.7EP

Vertex is $\left(2,37\right).$

Axis of symmetry is $x=2.$

y-intercept is $y=37.$

Explanation of Solution

Given:

The given equation is:

  $y=5x^2-20x+37$

Calculation:

The standard form of a quadratic function is:

  $y=a{x}^2+bx+c$

Consider the given equation and compare with standard form:

  $\begin{array}{l}a=5,b=-20,c=37\\ \text{x-coordinate of the vertex is:}\\ \frac{-b}{2a}\\ =-\frac{-20}{2\left(5\right)}\\ =\frac{20}{10}\\ =2\\ \text{y-coordinate of the vertex is:}\\ y=5{\left(0\right)}^2-20\left(0\right)+37\\ y=37\\ \therefore \text{ the vertex is }\left(2,37\right).\end{array}$

Now, calculating the axis of symmetry:

  $\begin{array}{l}\text{Axis of symmetry is given by: x=}\frac{-b}{2a}\\ \therefore \\ x=\frac{-b}{2a}\\ x=2\end{array}$

Calculating the y-intercept:

  $\begin{array}{l}y=5{\left(0\right)}^2-20\left(0\right)+37\\ y=37\end{array}$