what is the x-coordinate of the vertex of the parabola defined by the function f(x) = -9x^2+5x+4

y = -9x

^{2}+ 5x + 4There are three ways to find the x-coordinate of the vertex.

The easiest way is simply to remember that the x-coordinate of the vertex is always equal to -b/2a, where a is the coefficient of the x

^{2}term and b is the coefficient of the x term. In this case, a = -9 and b = 5, so x = -5/(2)(-9) =**5/18**---

The second way is to put the quadratic equation into the vertex form:

(y-k) = a(x-h)

^{2}Where (h,k) are the x and y coordinates of the vertex.

y = -9x

^{2}+ 5x + 4Subtract 4 from both sides, then factor out the coefficient of the x

^{2}termy - 4 = -9(x

^{2}+ (5/9)x)Now complete the square:

y - 4 - 9(5/18)

^{2}= -9(x^{2}- (5/9)x + (5/18)^{2}) (y - 4

^{25}/_{36})= -9(x - (5/18))^{2}So x = h =

**5/18**----

If you know calculus, the third way is to take the derivative of the quadratic and set it to zero:

dy/dx = -18x + 5

0 = -18x + 5

18x = 5

**x = 5/18**