Answer
Zeros: $-\dfrac{1}{3}$
$x$-intercepts: $-\dfrac{1}{3}$
Work Step by Step
To find the zeros of a function $f$, solve the equation $f(x)=0$
The zeros of the function are also the $x-$intercepts.
Let $f(x)=0$:
$$(3x+4)^2-6(3x+4)+9=0$$
Let $u=3x+4$, the original equation becomes
$$u^2-6u+9=0$$
By factoring
$$(u-3)(u-3) = 0$$
Use the Zero-Product Property by equating each unique factor to zero, then solve each equation to obtain:
\begin{align*}
u -3&= 0\\
u&=3
\end{align*}
To solve for $x$, we use $u=3x+4$
$$\because u = 3x+4$$
$$\therefore 3x = u-4$$
$$\therefore x = \dfrac{u-4}{3}$$
For $u=3$
$$x=\dfrac{3-4}{3} $$
$$x= -\dfrac{1}{3}$$
$\therefore x = -\dfrac{1}{3}$
