Determine the vertex of the graph of the quadratic function f(x)=x^2-5x+29/4
So that you will understand where the numbers come from, the given equation
is in the form y = ax^2 + bx + c
a = 1, b = -5, c = 29/4
The x coordinate of the vertex is (-b/2a); the y coordinate is (- (b^2 - 4ac)/4a)
x = (- (-5) / (2 * 1))
x = 5 / 2
x = 2.5
y = (- ((-5)^2 - (4 * 1 * (29/4))) / (4 * 1))
y = (- (25 - 29) / 4)
y = - (-4 / 4)
y = - (-1)
y = 1
The vertex is (2.5, 1)