The vertex of a parabolic function is located at (5,-4). One of its zeros (x-intercepts) occurs at x = 7. Where will its other zero (x-intercept) be located?

The answer to the question is

Its other zero (x-intercept) will be located at x = - 5

Step-by-step explanation:

To solve the question, we note that a parabolic function is of the form

ax² + bx + c = 0

Therefore we have the vertex occurring at the extremum where the slope = 0

or dy/dx = 2a+b = 0 also the x intercept occurs at x = 7, therefore when

ax² + bx + c = 0, x = 7 which is one of the solution

when x = 5, y = - 4

That is a*25 + 5*b + c = - 4 also

49*a + 7*b + c = 0

2*a + b = 0

Solving the system of equations we get

a = 0.2, b = - 0.4 and c = - 7

That is 0.2x² - 0.4x - 7 = 0 which gives

(x+5) (x-7) * 0.2 = 0

Therefore the x intercepts are 7 and - 5

the second intercept will be located at x = - 5