Answer
The zeros of f(x) are 2 and 6
Step-by-step explanation
The function:
[tex]f(x)=3x^2-24x+36[/tex]
has the form:
[tex]f(x)=ax^2+bx+c[/tex]
with the constants: a = 3, b = -24, and c = 36.
We can find the zeros of f(x) with the help of the quadratic formula, as follows:
[tex]\begin{gathered} x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ x_{1,2}=\frac{24\pm\sqrt{(-24)^2-4\cdot3\operatorname{\cdot}36}}{2\cdot3} \\ x_{1,2}=\frac{24\pm\sqrt{144}}{6} \\ x_1=\frac{24+12}{6}=6 \\ x_2=\frac{24-12}{6}=2 \end{gathered}[/tex]
