Find the vertex, the equation of the axis of symmetry, and the $y$-intercept of the given equation.

$y=-3x^2+6x-18$

A quadratic function, which is written in the form, $y=a{x}^2+bx+c$, where,$a\ne 0$ is called the standard form of the quadratic function.

For the function$y=a{x}^2+bx+c$

(1) If $a>0$, then the function has the minimum value at$x=-\frac{b}{2a}$ and the vertex is located at the minimum point.

(2) If $a<0$, then the function has the maximum value at$x=-\frac{b}{2a}$ and the vertex is located at the maximum point.

The axis of symmetry for the function$y=a{x}^2+bx+c$ is

$x=-\frac{b}{2a}$

The $y$-intercept of the function$y=a{x}^2+bx+c$ is always at $c$.

Compare the quadratic function $y=-3x^2+6x-18$with the standard quadratic function role="math" localid="1647850331775" $y=a{x}^2+bx+c$.

$a=-3,b=6,c=-18$

Substitute$a=-3$ and$b=6$ in $x=-\frac{b}{2a}$.

$$\begin{gathered} x=-\left(\frac{6}{2\left(-3\right)}\right) \\ x=-\left(\frac{6}{-6}\right) \\ x=-\left(-1\right) \\ x=1 \end{gathered}$$

So, the axis of symmetry is given by

$x=1$

Since, $a>0$.

So, the function has a minimum value at $x=1$.

Substitute$x=1$ in role="math" localid="1647850498284" $y=-3x^2+6x-18$.

$$\begin{gathered} y=-3{\left(1\right)}^2+6\left(1\right)-18 \\ y=-3\left(1\right)+6-18 \\ y=-3+6-18 \\ y=-21+6 \\ y=-15 \end{gathered}$$

So, the vertex point is located at the point $\left(1,-15\right)$.

Since, the $y$-intercept is given by $c$.

So, the $y$-intercept is $-18$.

Therefore the vertex is $\left(1,-15\right)$, the equation of the axis of symmetry is$x=1$ and the $y$-intercept is $-18$.