Question

find the minimum y-value on the graph of y=f(x).
f(x)=x^2+9x

Answer 1

take the derivative and set equal to 0
f'(x) = 2x + 9 = 0
2x = -9/2
x = -9/4
plug in to get y val
y = (-9/4)^2 + 9(-9/4)
y = 81/16 -81/4 = -243/16 = -15.1875

Answer 2

or use the equation to find vertex -b/2a
so -9/2(1) = -9/2 <- plug this in again to find y val

Answer 3

so is the anwer -9 or 0 ?

Answer 4

the answer is -15.1875

Answer 5

\[x ^{2} +9x = (x+\frac{9}{2})^{2}-(\frac{9}{2})^{2}\]

so the min is -81/4 = -20.25