Answer
Vertex: $(3,-12)$
Vertical intercept: $(0,15)$
Horizontal intercepts: $(1,0),(5,0)$
Work Step by Step
Given $$
h(x)=3 x^2-18 x+15.
$$ The parabola is in its standard form, $f(x) = ax^2+bx+c$. The values of the constants are: $$
\begin{aligned}
a&=3 \\
b&=-18 \\
c&=15.
\end{aligned}
$$ Find the $x$ and $y$ values of the vertex.
$$
\begin{aligned}
x&=\frac{-b}{2a} \\
x&=\frac{-(-18)}{2(3)} \\
x&=3\\
y&= 3\cdot (3)^2-18\cdot (3)+15\\
&= -12.
\end{aligned}
$$ The vertex is $(3,-12)$.
The horizontal intercept is found by setting the $y$ component of the parabola to zero and solving for $x$.
$$
\begin{aligned}
3 x^2-18 x+15&=0 \\
\frac{3x^2-18 x+15}{3}&= 0\\
x^2-6 x+5&= 0 \\
x^2-5 x-x+5&= 0 \\
x(x-5)-1(x-5)&= 0\\
(x-5)(x-1)&= 0.
\end{aligned} $$
This gives: $$
\begin{aligned}
x-5&=0 \\
x&=5 \\
x-1&=0 \\
x&=1.
\end{aligned}
$$ The horizontal intercepts are $(1,0)$ and $(5,0)$.
The graph has a horizontal component for $x= 0$ and $y = c = 15$. The vertical intercept is $(0,15)$.
