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

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