For each parabola, find (a) which ways it opens, (b) the axis of symmetry, (c) the vertex, (d) the x-and y-intercepts, and (e) the maximum or minimum value.

$y=-3x^2+12x-8$

The given function is$y=-3x^2+12x-8$

Compare the given equation with the standard form $y=a{x}^2+bx+c$we get$a=-3,b=12,andc=-8$

Since the value of a is negative, so the parabola opens down.

The axis of symmetry is the vertical line given by:

$$\begin{gathered} x=-\frac{b}{2a} \\ =-\frac{12}{2(-3)} \\ =2 \end{gathered}$$

The axis of symmetry is $x=2$.

Since the vertex is a point on the line of symmetry so the x-coordinate of the vertex is 2.

Now determine the respective y-coordinate.

$$\begin{gathered} y=-{3\left(2\right)}^2+12\left(2\right)-8 \\ =-12+24-8 \\ =4 \end{gathered}$$

The vertex is$(2,4)$.

For y-intercept put $x=0$into the given equation.

$$\begin{gathered} y=-3{\left(0\right)}^2+12\left(0\right)-8 \\ =-8 \end{gathered}$$

The y-intercept is point $(0,-8)$.

For x-intercept put$y=0$ into the given equation.

role="math" localid="1653829355415" $$\begin{gathered} -3x^2+12x-8=0 \\ x=\frac{-12\pm \sqrt{{12}^2-4(-3)(-8)}}{2(-3)} \\ =\frac{-6\pm 2\sqrt{3}}{-3} \end{gathered}$$

The x-intercepts are $\left(\frac{-6-2\sqrt{3}}{-3},0\right)and\left(\frac{-6+2\sqrt{3}}{-3},0\right)$.

Since the parabola opens down, the graph will have the maximum point at the vertex.

The maximum value is 4 at $x=2$.