Chapter 9.2, Problem 8PPE

To determine

The equation of axis of symmetry and the coordinates of the vertex of the graph of each function.

Answer to Problem 8PPE

Equation of axis of symmetry: $x=2$

  $\text{Vertex is }\left(2,13\right)$

  

Explanation of Solution

Given:

The given function is:

  $y=-3x^2+12x+1$

Calculation:

Compare the given function with $y=a{x}^2+bx+c$ .

  $\begin{array}{l}here;\\ a=-3\\ b=12\\ c=1\end{array}$

  $\begin{array}{l}\text{Equation of axis of symmetry is:}\\ x=-\frac{b}{2a}\end{array}$

Therefore, the axis of symmetry is:

  $\begin{array}{l}x=-\frac{b}{2a}\\ x=-\frac{-12}{2\left(-3\right)}\\ x=2\end{array}$

The x-coordinate of the given function is $x=2$ . Now, the y-coordinate will be:

  $\begin{array}{l}y=-3x^2+12x+1\\ y=-3{\left(2\right)}^2+12\left(2\right)+1\\ y=-12+24+1\\ y=13\\ So,\text{Vertex is }\left(2,13\right)\end{array}$

Finding the other two points on the graph of the given function:

  $\begin{array}{l}\text{When x = 1;}\\ \text{y = -3}{\left(1\right)}^2+12\left(1\right)+1\\ y=10\\ \text{Point is }\left(1,10\right).\\ \text{When x = 0;}\\ \text{y = -3}{\left(0\right)}^2+12\left(0\right)+1\\ y=1\\ \text{Point is }\left(0,1\right).\end{array}$

Now, graph the vertex (2,13) on the graph and with the other two points (1,10) and (0,1).

Reflect the points across the axis of symmetry to get two more points on the graph.

Finally, connect these points to the parabola.

The graph will be: