Respuesta :

at the roots, y=0

[tex]\begin{gathered} \Rightarrow x^2+8x+12=0 \\ \Rightarrow(x+4)^2-(4)^2+12=0 \\ \Rightarrow(x+4)^2=4 \\ \Rightarrow x+4=\pm2 \\ \Rightarrow x=-4\pm2 \\ \Rightarrow x=-4+2\text{ or -4-2} \\ \Rightarrow x=-2\text{ or -6} \\ \end{gathered}[/tex]

So two of the five points are:

(-2,0) and (-6,0)

At the vertex, dy/dx = 0

[tex]\begin{gathered} \frac{dy}{dx}=2x+8 \\ \text{When }\frac{dy}{dx}=0 \\ 2x+8=0 \\ \Rightarrow x=-4 \\ y=-4 \\ (-4,-4)\text{ is the vertex} \end{gathered}[/tex]

When x=-1, y=5

(-1, 5)

When x=-3, y=-3

(-3,-3)

So the five points are

when x= -3, y=-3

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