Answer
Vertex: $(2,-2)$
Vertical intercept: $(0,4)$
Horizontal intercepts: $(3.15,0),(0.85,0)$
Work Step by Step
Given $$
f(x)=1.5 x^2-6 x+4.
$$ The parabola is in its standard form, $f(x) = ax^2+bx+c$. The values of the constants are: $$
\begin{aligned}
a&=1.5 \\
b&=-6 \\
c&=4.
\end{aligned}
$$ Find the $x$ and $y$ values of the vertex.
$$
\begin{aligned}
x & =\frac{-b}{2 a} \\
& =\frac{-(-6)}{2(1.5)} \\
& =2
\end{aligned}
$$ $$
\begin{aligned}
y & =1.5 \cdot 2^2-6 \cdot 2+4 \\
& =-2
\end{aligned}
$$ $$
\text { Vertex: }(2,-2)
$$ Find the vertical intercept.
$$
\begin{aligned}
&y=c=4\\
&\text { Vertical intercept: }(0,4)
\end{aligned}
$$ Find the horizontal intercept by setting $y= 0$ to find the $x$ values.
$$
\begin{aligned}
\frac{1.5 x^2-6 x+4}{1.5} & =\frac{0}{1.5} \\
x^2-4 x+\frac{40}{15} & =0 \\
x^2-4 x+2^2 & =\frac{-40}{15}+2^2 \\
(x-2)^2 & =\frac{4\cdot15-40}{15} \\
(x-2)^2 & =\frac{20}{15}\\
(x-2)^2&=\frac{20}{15}\\
(x-2)^2 & = \pm \sqrt{\frac{20}{15}} \\
x & =2 \pm \sqrt{\frac{20}{15}}\\
\end{aligned}
$$
$$
\begin{aligned}
x & =2+\sqrt{\frac{20}{15}} & x & =2-\sqrt{\frac{20}{15}} \\
& \approx3.15 & & \approx0.85
\end{aligned}
$$ $$
\text {Vertical intercept:s }(3.15,0),(0.85,0)
$$
