## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$a=-1$ or $a=-5$
If a point is on the graph, the $(x,y)$ coordinates satisfy the equation. Here, $(a,-5)$ is on the graph, therefore we should plug in these values into the given equation to obtain: $y=x^2+6x$ $-5=a^2+6a$ $0=a^2+6a+5$ By using the quadratic formula: $a_{1,2}=\dfrac{-6\pm\sqrt{6^2-4(1)(5)}}{2(1)}=\dfrac{-6\pm\sqrt{36-20}}{2}=\dfrac{-6\pm4}{2}$ $a_1=\dfrac{-6+4}{2}=-1$ $a_2=\dfrac{-6-4}{2}=-5$ The points $(-1,-5)$ and $(-5,-5)$ are on the graph.