Answer
Vertex: $(-6.67,-48.33)$
Vertical intercept: $(0,-15)$
Horizontal intercepts: $(-14.69,0),(1.36,0)$
Work Step by Step
Given $$
g(x)=0.75 x^2+10 x-15.
$$ The parabola is in its standard form, $f(x) = ax^2+bx+c$. The values of the constants are: $$
\begin{aligned}
a&=0.75 \\
b&=10 \\
c&=-15.
\end{aligned}
$$ Find the $x$ and $y$ coordinates of the vertex.
$$\begin{align*}
x &=\frac{-b}{2 a}=\frac{-10}{2(0.75)}\approx-6.667\\
y & \approx 0.75(-6.667)^2+10(-6.667)-15 \\
& \approx -48.33
\end{align*}
$$ $$
\text { Vertex: }(-6.67,-48.33).
$$ Find the vertical intercept.
$$
\begin{aligned}
&y=c=-15\\
&\text { Vertical intercept: }(0,-15).
\end{aligned}
$$ Find the horizontal intercept by setting $y= 0$ to find the $x$ values.
$$
\begin{aligned}
\frac{0.75 x^2+10 x-15}{0.75}&=\frac{0}{0.75} \\
x^2+\frac{1000}{75} x-20&=0 \\
x^2+\frac{40}{3} x+\left(\frac{40}{6}\right)^2&=20+\left(\frac{40}{6}\right)^2\\
\left(x+\frac{40}{6}\right)^2& =\frac{36 \cdot 20}{36}+\frac{1600}{36} \\
\left(x+\frac{40}{6}\right)^2&=\frac{720+1600}{36} \\
\left(x+\frac{40}{6}\right)^2&=\frac{2320}{36}\\
\left(x+\frac{40}{6}\right)&= \pm \frac{\sqrt{2320}}{6}\\
x&=-\frac{40}{6} \pm \frac{\sqrt{2320}}{6}
\end{aligned}
$$ The solutions are: $$
\begin{aligned}
x & =-\frac{40-\sqrt{2320}}{6} \\
& \approx 1.36\\
x & =-\frac{40+\sqrt{2320}}{6} \\
& \approx -14.69.
\end{aligned}
$$ $$
\text { Horizontal intercepts: }(-14.69,0),(1.36,0).
$$
