What is the y-value of the vertex of the function f(x) = –(x + 8)(x – 14)? the y-value of the vertex is .

      Best   y=18x^2+9x+14

y=18(x^2+(1/2)x + 1/16) + 14 - 9/8

y = 18(x + 1/4)^2 + 103/8

For y = a*(x - h)^2 + k, (h, k) the vertex

Your vertex: (- 1/4, 103/8)

Equation of axis of symmetry is x = (x-coordinate of vertex) OR x = - 1/4

y-intercept is y-value when x = 0, y = 14

x-intercept(s) do not exist for this upward opening parabola whose vertex y-value is above the x-axis.

The minimum of the function is the y-value of the vertex, 103/8.
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Seems like Dizzle was in too much of a hurry and Jeff just copied Dizzle's answer.

The correct way of doing what dizzle TRIED to do:

For y = a*x^2 + b*x + c, the vertex occurs when x = - b / (2*a)

y=18x^2+9x+14

a = 18
b = 9

- b / (2*a) = - (9) / (2*18) = - 9 / 36 = - 1/4

Take that value for x, evaluate function at that value to get y.
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Was so giddy about dizzle's faux pas that I originally did this work. May as well share it:

x-intercept(s) can be found by setting function equal to zero and solving:

y = 18(x + 1/4)^2 + 103/8

0 = 18(x + 1/4)^2 + 103/8

18(x + 1/4)^2 = - 103/8

(x + 1/4)^2 = - 103/144

x + 1/4 = +/- i*sqrt(103)/12

x = - 1/4 +/- i*sqrt(103)/12 OR x = [- 3 +/- i*sqrt(103)] / 12

These are the roots of the equation f(x) = 0

Since the roots are complex, there are no x-intercepts. The function is entirely above the x-axis.