how do I find the vertex of y=2(x+2)^2+4

Question

asnaseer · Answer

dy/dx is zero at the vertex.
so calculate what dy/dx is, set it equal to zero, that should give you a value for x, then plug this value of x into the original equation to get the y value.

anonymous · Answer

Is it (-5, 4)

asnaseer · Answer

no
what did you get for dy/dx?

anonymous · Answer

I just graphed it

asnaseer · Answer

have you learnt how to do dy/dx?
or are you supposed to graph this and find the answer by inspection?

anonymous · Answer

i have not seen the dy/dx before

anonymous · Answer

yes because I also am suppose to find the intercepts

asnaseer · Answer

if you just want to check whether or not you graphed it correctly, just use wolfram alpha: http://www2.wolframalpha.com/input/?i=y%3D2%28x%2B2%29%5E2%2B4

anonymous · Answer

When you have a parabola in the form y=a(x-h)^2+k, the vertex is (h,k)

anonymous · Answer

yea that is what I got.

asnaseer · Answer

@Danyel - h is NOT equal to -5. look very carefully at what @Xavier said.

anonymous · Answer

I re did it and got (-2, 4)

asnaseer · Answer

ok - that is correct :-)

anonymous · Answer

do I take these nimber to find the intercepts

anonymous · Answer

numbers

asnaseer · Answer

when the graph intercepts the x-axes, the y value is zero
when it intercepts the y-axes, the x value is zero

anonymous · Answer

ok