Answer
See the graph
Work Step by Step
Given: $y=-\frac{1}{3}x^2-5x+2$
The coefficients are $a =-\frac{1}{3}$, $b =-5$, and $c=2$. Because $a \lt 0$, the parabola opens down.
Find the vertex.
$x=-\frac{b}{2a}=\frac{-(-5)}{2.(-\frac{1}{3})}=-\frac{15}{2}$
Then find the y-coordinate of the vertex.
$y=-\frac{1}{3}(-\frac{15}{2})^2-5(-\frac{15}{2})+2=\frac{83}{4}$
Draw the axis of symmetry $x =-\frac{15}{2}$
The y-intercept is $2$. Plot the point $(0, 2)$. Then reflect this point in the axis of symmetry to plot another point, $(-15,2)$.
Evaluate the function for another value of $x$, such as $x =-3$.
$y=-\frac{1}{3}(-3)^2-5(-3)+2=14$
Plot the point $(-3, 14)$ and its reflection $(-12, 14)$ in the axis of symmetry.
Draw a parabola through the plotted points.
