Chapter 1.1, Problem 28E

Solution

Graph the function. Label the vertex and axis of symmetry.

Axis of symmetry (x) $=\frac{-b}{2a}=-2.66$

Vertex $=(-2.66,4.33)$

Given:

The given function $y=\frac{-3}{4}{x}^2-4x-1$

Concept Used:

• The vertex is the highest point if the parabola opens downward and the lowest point if the parabola opens upward.
• The axis of symmetry is the line that cuts the parabola into 2 matching halves and the vertex lies on the axis of symmetry.

Calculation:

The graph of the given function $y=\frac{-3}{4}{x}^2-4x-1$ is -

  

The given function $y=\frac{-3}{4}{x}^2-4x-1$

Here $a=\frac{-3}{4},b=-4,c=-1$

Axis of symmetry (x) $=\frac{-b}{2a}$

  $\begin{array}{l}=\frac{-(-4)}{2(\frac{-3}{4})}\\ =\frac{4\cdot 4}{-6}\\ =\frac{-8}{3}\\ =-2.66\end{array}$

So Axis of symmetry (x) $=\frac{-b}{2a}=-2.66$

Vertex $=\left(\frac{-b}{2a},y(\frac{-b}{2a})\right)$

  $\begin{array}{c}y\left(\frac{-b}{2a}\right)=y\left(\frac{-8}{3}\right)\\ y=\frac{-3}{4}{x}^2-4x-1\\ =\frac{-3}{4}{\left(\frac{-8}{3}\right)}^2-4\left(\frac{-8}{3}\right)-1\\ =\frac{-3}{4}\left(\frac{64}{9}\right)+\frac{32}{3}-1\\ =\frac{-16}{3}+\frac{32}{3}-1\\ =\frac{16}{3}-1\\ =\frac{16-3}{3}\\ =\frac{13}{3}\\ =4.33\end{array}$

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

Vertex $=(-2.66,4.33)$