What is the vertex of f(x)=(x-3)^2-4

Question

I know that I need to use the formula -b/2a, I'm just unsure why my answers don't make sends.

Answer 1

f(x)= (x-3)^2-4

 

take the derivative and set=o, this is the x-coordinate of the vertex

 

df/dx= 2*(x-3)=0

 

that says, x-3=0 or x=3

 

plug in x into the original equation, we get

 

f(x)=(0-0)^2-4= -4

 

hence vertex is (3,-4)

Answer 2

f(x) = (x-3)² - 4

 

This is the "Vertex" form of a quadratic equation:

 

f(x) = (x-h)² + k

 

where (h,k) is the location of the vertex.  In your case. h=3 and k=-4, so the vertex is located at the point (3,-4).

Answer 3

f(x) = (x - h)² - k, the vertex is (h, k)

f(x) = (x - 3)² - 4, the vertex is (3, -4)