Answer
Zero: $\dfrac{1}{3}$
$x$-intercept: $\dfrac{1}{3}$
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 $P(x)=0$:
$$9x^2-6x+1=0$$
Comparing $9x^2-6x+1=0$ to $ax^2+bx+c=0$ to find $a,b \text{ and } c$
$$\therefore a = 9, b=-6 , c =1$$
Evaluating the discriminant $b^2-4ac$
$$b^2-4ac = (-6)^2-4 \times 9 \times 1 = 0$$
Since the discriminant is equal to zero, then there is a real repeated root.
The quadratic formula is given by:
$$x= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$$
$$x= \dfrac{6\pm \sqrt{0}}{2\times 9}$$
$$x=\dfrac{6\pm 0}{18}$$
$\therefore x =\dfrac{1}{3}$
