Chapter 8.8, Problem 27WE

To determine

To find: Thecoordinates of the vertex, the equation of the axis of symmetry and the graph of the function.

Answer to Problem 27WE

The coordinate of the vertex is $\left(-\frac{5}{2},\frac{49}{4}\right)$ and the equation of the symmetry is $x=-\frac{5}{2}$ .

Explanation of Solution

Given information:

The given equation is,

  $y=-x^2-5x+6$

The x -coordinate of the vertex is calculated as,

  $\begin{array}{c}x=-\frac{b\left(\text{coefficent of }x\right)}{2a\left(\text{coefficent of }{x}^2\right)}\\ =-\frac{\left(-5\right)}{2\left(-1\right)}\\ =-\frac{5}{2}\end{array}$

The y -coordinate of the vertex is calculated as,

  $\begin{array}{l}y=-x^2-5x+6\\ y=-{\left(-\frac{5}{2}\right)}^2-5\left(-\frac{5}{2}\right)+6\\ y=\frac{49}{4}\end{array}$

The coordinate of the vertex is $\left(-\frac{5}{2},\frac{49}{4}\right)$ .

The equation of axis od symmetry is calculated as,

  $\begin{array}{c}x=-\frac{b\left(\text{coefficent of }x\right)}{2a\left(\text{coefficent of }{x}^2\right)}\\ =-\frac{\left(-5\right)}{2\left(-1\right)}\\ x=-\frac{5}{2}\end{array}$

Consider the graph of the given equation,

  

Thus, the coordinate of the vertex is $\left(-\frac{5}{2},\frac{49}{4}\right)$ and the equation of the symmetry is $x=-\frac{5}{2}$ .