Answer
Vertex: $(-1.2, 2.8)$
Vertical intercept: $(0,10)$
Horizontal intercepts: None
Work Step by Step
Given $$
g(x)=5 x^2+12 x+10.
$$ The parabola is in its standard form, $f(x) = ax^2+bx+c$. The values of the constants are: $$
\begin{aligned}
a&=5 \\
b&=12 \\
c&=10.
\end{aligned}
$$ Find the $x$ and $y$ values of the vertex.
$$
\begin{aligned}
x&=\frac{-b}{2a} \\
x&=\frac{-(12)}{2(5)} \\
x&=-1.2\\
y&= 5\cdot (-1.2)^2+12\cdot (-1.2)+10\\
&= 2.8.
\end{aligned}
$$ The vertex is $(-1.2,2.8)$.
The horizontal intercept is found by setting the $y$ component of the parabola to zero and solving for $x$. $$
\begin{aligned}
5x^2+12x+10&=0.
\end{aligned}
$$ $$
\begin{aligned}
& x=\frac{-12 \pm \sqrt{12^2-4 \cdot 5 \cdot 10}}{2 \cdot 5} \\
& x=\frac{-12 \pm \sqrt{-56}}{10}.
\end{aligned}
$$ There is no horizontal intercept since we cannot take the square root of a negative number.
The graph has a vertical intercept for $x= 0$ and $y = c = 10$. The vertical intercept is $(0,10)$.
