Question 536670
IF YOU CAN FACTOR (using the quadratic formula to help is allowed)
{{{x^2-14x+48=(x-8)(x-6)}}} so it will be positive for x<6 and for x>8, zero at x=6 and x=8 , and negative in between.
IF YOU WILL NOT FACTOR
The coefficient of the {{{x^2}}} term is positive (+1), so the parabola opens up (smiling shape). A parabola with that shape might be negative somewhere in the middle, but will definitely have positive y values at both ends. That eliminates answers C and D.
For x=10 {{{x^2-14x+48=10^2-14*10+48=100-140+48=8}}}
and that eliminates answer A.