Chapter 7.4, Problem 84STP

To determine

Write the equation $y=-\frac{1}{3}{x}^2+8x$ in vertex form if not already in that form and then identify vertex, axis of symmetry, direction of opening and sketch the graph of the function.

Answer to Problem 84STP

The vertex form of equation is $y=-\frac{1}{3}{\left(x-12\right)}^2+48$ , vertex is $\left(12,48\right)$ , axis of symmetry is $x=12$ and direction of opening of graph is downwards.

Explanation of Solution

Given information: The given equation is,

  $y=-\frac{1}{3}{x}^2+8x\mathrm{......}\left(1\right)$

Calculation: The equation (1) in vertex form is written as,

  $\begin{array}{l}y=-\frac{1}{3}{x}^2+8x\mathrm{......}\left(\text{Given}\right)\\ y=-\frac{1}{3}\left(x^2-24x\right)\\ y+\left(-48\right)=-\frac{1}{3}\left(x^2-2\left(x\right)\left(12\right)+{\left(12\right)}^2\right)\mathrm{....}\left(\text{Complete square}\right)\\ y=-\frac{1}{3}{\left(x-12\right)}^2+48\mathrm{....}\left(2\right)\end{array}$

Now the equation (2) is in vertex form. Therefore, comparing (2) with vertex form $y=a{\left(x-h\right)}^2+k$ of a standard parabola, the vertex is $\left(12,48\right)$ .

Now the vertex is the point $\left(12,48\right)$ , therefore the equation for axis of symmetry is the line passing through the $\left(x\right)$ coordinate of the vertex. Therefore the equation for axis of symmetry is $x=12$ .

From the vertex form of parabola the leading coefficient of ${\left(x-12\right)}^2$ is $-\frac{1}{3}$ which is negative therefore the parabola will be downward parabola which will open downwards.

Graph: The graph for the function is shown as,