Chapter 2.2, Problem 30E

To determine

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

Answer to Problem 30E

The vertex of the function $y=-\frac{5}{2}{x}^2-4x-1$ is $(-0.8,0.6)$ .

Axis of symmetry $x=-0.8$ .

  

Explanation of Solution

Given information: $y=-\frac{5}{2}{x}^2-4x-1$ .

Identifying the vertex:

General form of a quadratic function $y=a{x}^2+bx+c$ ,

From the function $y=-\frac{5}{2}{x}^2-4x-1$ ,

  $a=-\frac{5}{2},b=-4,c=-1$

The $x-\text{coordinate}$ of a quadratic function is $x=-\frac{b}{2a}$ ,

  $\begin{array}{l}x=-\frac{(-4)}{2\times (-\frac{5}{2})}\\ x=-\frac{(-4)}{(-5)}\\ x=-0.8\end{array}$

Substituting $x=-0.8\text{in}y=-\frac{5}{2}{x}^2-4x-1$ ,

  $\begin{array}{l}$ y=-\frac{5}{2}{(0.8)}^2-4(-0.8)-1$y=-\frac{5}{2}(0.64)+3.2-1\\ y=-1.6+3.2-1\\ y=0.6\end{array}$

The vertex of the given quadratic function $(h,k)$ is $(-0.8,0.6)$ .

The axis of symmetry is a vertical line which passes through the vertex of the parabola.

The $x-\text{coordinate}$ of the vertex is the equation of axis of symmetry $x=-0.8$ .