91Ó°ÊÓ

We have:

$f\left(x\right)=-2x^2+8x+4$

x-intercept(s): It is the point where $y=0.$It follows:

$-2x^2+8x+4=0$

For a quadratic equation of the form $a{x}^2+bx+c=0$, the solutions are:

$$\begin{gathered} x_{1, 2}=\frac{-bÂ\pm \sqrt{b^2-4ac}}{2a} \\ \mathrm{For} a=-2, b=8, c=4 \\ {x}_{1, 2}=\frac{-8Â\pm \sqrt{8^2-4\left(-2\right)· 4}}{2\left(-2\right)} \\ {x}_{1, 2}=\frac{-8Â\pm  4\sqrt{6}}{2\left(-2\right)} \\ x=2-\sqrt{6}, x=2+\sqrt{6} \end{gathered}$$

y-intercept(s): It is the point where $y=0$, it follows:

$$\begin{gathered} y=-2{\left(0\right)}^2+8\left(0\right)+4 \\ y=4 \end{gathered}$$

Therefore, the intercepts are:

$\mathrm{X} \mathrm{Intercepts}: \left(2-\sqrt{6}, 0\right), \left(2+\sqrt{6}, 0\right), \mathrm{Y} \mathrm{Intercepts}: \left(0, 4\right)$

$\mathrm{The} \mathrm{vertex} \mathrm{of} \mathrm{an} \mathrm{up}-\mathrm{down} \mathrm{facing} \mathrm{parabola} \mathrm{of} \mathrm{the} \mathrm{form} y=a{x}^2+bx+c \mathrm{is} x_v=-\frac{b}{2a}$

The parabola params are:

$a=-2, b=8, c=4$

$$\begin{gathered} x_v=-\frac{b}{2a} \\ {x}_v=-\frac{8}{2\left(-2\right)} \\ {x}_v=2 \end{gathered}$$

Plug $x_v=2$into original equation to find $y_v:$

$y_v=12$

Therefore, the vertex is$\left(2, 12\right)$.

The axis of symmetry of a parabola is a vertical line that divides the parabola into two congruent halves.

$f\left(x\right)=-2x^2+8x+4$

So, the line $x=2$is the axis of symmetry.

Use intercepts, vertex, and the equation of axis of symmetry to draw the graph: