Answer
$y= 6 x^2-\frac{21}{10} x-\frac{9}{5}$
Work Step by Step
Since the zeros of the function are located at $x=- \frac{2}{5}$ and $x= \frac{3}{4}$, we can write $$
\begin{aligned}
y & =a\left(x+\frac{2}{5}\right)\left(x-\frac{3}{4}\right) \\
& =a\left[\left(x+\frac{2}{5}\right) x-\frac{3}{4}\left(x+\frac{2}{5}\right)\right] \\
& =a\left(x^2+\frac{2}{5} x-\frac{3}{4} x-\frac{3}{10}\right) \\
& =a\left(x^2+\frac{2\cdot4-3\cdot5}{20} x-\frac{3}{10}\right)\\
&= a\left(x^2-\frac{7}{20} x-\frac{3}{10}\right)\\
y&= a x^2-\frac{7}{20} a x-\frac{3}{10} a.
\end{aligned}
$$ Given that $f(2)=18$, we can use the point $(x,y) = (2 , 18) $ to find the constant $a$.
$$
\begin{aligned}
a\left(2+\frac{2}{5}\right)\left(2-\frac{3}{4}\right) & =18 \\
a\left(\frac{2 \cdot 5+2}{5}\right)\left(\frac{2 \cdot 4-3}{4}\right) & =18 \\
a\left(\frac{12}{5}\right)\left(\frac{5}{4}\right) & =18 \\
3 a & =18 \\
a & =6.
\end{aligned}
$$This gives $$
\begin{aligned}
y & =6 x^2-\frac{7}{20}(6) x-\frac{3}{10}(6) \\
& =6 x^2-\frac{21}{10} x-\frac{9}{5}.
\end{aligned}
$$
