Given the function:
[tex]-3x^2-12x-10[/tex]In order to graph the quadratic function, we find some points on the parabola.
Vertex
Using the vertex formula:
[tex]\begin{gathered} \text{Vertex}=(-\frac{b}{2a},\frac{4ac-b^2}{4a}) \\ =(-\frac{-12}{2\times-3},\frac{4(-3)(-10)-(-12)^2}{4\times-3}) \\ =(-2,2) \end{gathered}[/tex]y-intercept
When x=0
[tex]\begin{gathered} y=-3(0)^2-12(0)-10=-10 \\ \implies(0,-10) \end{gathered}[/tex]Next, look for one point to the right of the vertex and one to the left.
When x=-1
[tex]\begin{gathered} y=-3(-1)^2-12(-1)-10=-3+12-10=-1 \\ \implies(-1,-1) \end{gathered}[/tex]Similarly, when x=-3
[tex]\begin{gathered} y=-3(-3)^2-12(-3)-10=-27+36-10=-1 \\ \implies(-3,-1) \end{gathered}[/tex]Similarly, when x=-4
[tex]\begin{gathered} y=-3(-4)^2-12(-4)-10=-48+48-10=-10 \\ \implies(-4,-10) \end{gathered}[/tex]The graph showing the points (-2,2), (0,-10), (-1,-1), (-3,-1) and (-4,-10) is attached below:
