Chapter 12.4, Problem 10MRE

To determine

To find: the vertex and the axis of symmetry of the graph of the given equation.

Answer to Problem 10MRE

The coordinate of the vertex is: $\left(4\frac{1}{2},-\frac{93}{4}\right)$ .

The axis of symmetry is $x=4\frac{1}{2}$ .

Explanation of Solution

Given information:

Given equation,

  $y=-x^2+9x-3$

Concept used:

Vertex of parabola of the form $y=f\left(x\right)=a{x}^2+bx+c$ is given by

  $\left(-\frac{b}{2a},f\left(-\frac{b}{2a}\right)\right)$

The equation of axis of symmetry:

  $x=-\frac{b}{2a}$

Calculation:

Consider the given equation:

  $y=-x^2+9x-3$

  $x$ Coordinate of the vertex:

  $\begin{array}{c}x=-\frac{9}{2\left(-1\right)}\\ =\frac{9}{2}\\ =4\frac{1}{2}\end{array}$

To find out the $y$ coordinate, put $x=\frac{9}{2}$ in the given function:

  $\begin{array}{c}y=-x^2+9x-3\\ =-{\left(\frac{9}{2}\right)}^2+9\left(\frac{9}{2}\right)-3\\ =-\frac{81}{4}+\frac{81}{2}-3\\ =-\frac{93}{4}\end{array}$

Hence, the coordinate of the vertex is: $\left(4\frac{1}{2},-\frac{93}{4}\right)$ .

The equation of axis of symmetry:

  $\begin{array}{c}x=-\frac{9}{2\left(-1\right)}\\ =\frac{9}{2}\\ =4\frac{1}{2}\end{array}$

Hence, the axis of symmetry is $x=4\frac{1}{2}$ .