The function h(x) is quadratic and h(3) = h(–10) = 0. which could represent h(x)?

h(x) = x2 – 13x – 30

h(x) = x2 – 7x – 30

h(x) = 2x2 + 26x – 60

h(x) = 2x2 + 14x – 60

Option D.

Step-by-step explanation:

We can solve this question by finding the values of h(3) and h(-10) from each option given.

Option A.

h(x) = x² - 13x - 30

h(3) = 3² - (13×3) - 30

      = 9 - 39 - 30

      = -60

h(-10) = (-10)² - 13(-10) - 30

         = 100 + 130 - 30

         = 200

Therefore, h(3) ≠ h(-10)

Option B.

h(x) = x² - 7x - 30

h(3) = 3² - 7×3 - 30

      = 9 - 21 - 30

      = -42

h(-10) = (-10)² - 7(-10) - 30

         = 100 + 70 - 30

         = 140

h(3) ≠ h(-10)

Option C.

h(x) = 2x² + 26x - 60

h(3) = 2(3)² + 26×3 - 60

      = 18 + 78 - 60

      = 36

h(-10) = 2(-10)² + 26(-10) - 60

         = 200 - 260 - 60

         = -120

h(3) ≠ h(-10)

Option D.

h(x) = 2x² + 14x - 60

h(3) = 2(3)² + 14(3) - 60

      = 18 + 42 - 60

      = 0

h(-10) = 2(-10)² + 14(-10) - 60

         = 200 - 140 - 60

         = 0

h(3) = h(-10) = 0

Therefore, option D is the answer.

answer: on graphs, the open and closed circles, or vertical asymptotes drawn as dashed lines us identify discontinuities.

as before, graphs and tables allow us to estimate at best.

when working with formulas, getting zero in the denominator indicates a point of discontinuity.

when working with piecewise-defined functions, check for discontinuities at the transition points where one piece ends and the next begins.

step-by-step explanation: