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

x=-2

infinite number of lines, so there is an infinite number of equations.

step-by-step explanation:

there are an infinite number of line that passes through the point (-2,5)

x=-2   is one line that passes through the point

y = 5 is another line that passes through the point

(these two lines are perpendicular to each other).

since the number of   numbers are infinite and the slope is the ratio of two numbers,(change in y over change in x) we can have an infinite number of slopes.     our only condition is that it passes through the point (-2,5)