Question

find the minimum y-value on the graph of y=f(x)
f(x)=x^2+4x-1

Answer 1

Put it in vertex form by completing the square.

Answer 2

\[
x^2+bx = \left(x+\frac b 2\right)^2-\left(\frac b 2\right)^2
\]
So \[
x^2+4x=(x+2)^2-4
\]And \[ y = ((x+2)^2-4)-1 = (x+2)^2-5\]

Answer 3

This finally leaves us at \[
y-(-5)=(x-(-2))^2
\]So the vertex is at \((-2,-5)\).

Answer 4

The vertex must be the lowest point.

Answer 5

@uniquejamaica get it?

Answer 6

yes thank you