Answer:
Step-by-step explanation:
When the question comes with answer choices, please share them. In this case it sounds like there were four graphs from which to choose.
f(x) = |x^2 − x − 2| is always zero or greater, due to the absolute value function.
x^2 − x − 2 = 0 factors to (x - 2)(x + 1) = 0, so the zeros of f(x) are {-1, 2}.
Plot x-intercepts {-1, 2}. The axis of symmetry is the vertical line x = 1/2, which is precisely halfway between the x-intercepts. If we now choose the test number x = 0, we find the value of f(0) to be |-2}, which tells us that the y-intercept is (0, |-2|), or (0, 2), so we have a parabolic curve opening down between x = -1 and x = 2 and touching (but not crossing) the x-axis at those x-values. To the left of x = -1 the curve increases steadily from y = 0 in Quadrant II; to the right of x = 2, the curve increases steadily from y = 0 in Quadrant I.
