[tex]\begin{gathered} \text{Given} \\ f(x)=-2x^2-8x+1 \end{gathered}[/tex]
First, find the vertex of the given function, we have
[tex]\begin{gathered} f(x)=-2x^2-8x+1 \\ \\ \text{The coefficients are } \\ a=-2,b=-8,c=1 \\ \\ \text{The x-coordinate of the vertex is at } \\ x=-\frac{b}{2a} \\ x=-\frac{-8}{2(-2)} \\ x=-\frac{-8}{-4} \\ x=-2 \end{gathered}[/tex]
Next, substitute x = -2. to the given function and we get
[tex]\begin{gathered} f(x)=-2x^{2}-8x+1 \\ f(-2)=-2(-2)^2-8(-2)+1 \\ f(-2)=-2(4)+16+1 \\ f(-2)=-8+17 \\ f(-2)=9 \end{gathered}[/tex]
Therefore, the vertex is at (-2.9).
The axis of symmetry is at x = -2.
The y-intercept at y = 1.
The x-intercepts are the following:
[tex]\begin{gathered} x=\frac{ -b \pm\sqrt{b^2 - 4ac}}{ 2a } \\ x = \frac{ -(-8) \pm \sqrt{(-8)^2 - 4(-2)(1)}}{ 2(-2) } \\ x=\frac{8\pm\sqrt{64-(-8)}}{-4} \\ x = \frac{ 8 \pm \sqrt{72}}{ -4 } \\ x = \frac{ 8 \pm 6\sqrt{2}\, }{ -4 } \\ \text{ Which becomes} \\ \\ x=\frac{8+6\sqrt{2}\,}{-4}\approx−4.12132 \\ x=\frac{8-6\sqrt{2}\,}{-4}\approx0.12132 \end{gathered}[/tex]
Graphing the function we get
