What is the vertex of #y= -(x+1)^2 +17 #?

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Answer 1 · 2015-12-16T05:00:43

vertex#=(-1,17)#

The general equation of a quadratic equation in vertex form is:

#y=a(x-h)^2+k#

where:
#a=#vertical stretch/compression
#h=#x-coordinate of vertex
#k=#y-coordinate of vertex

Looking back at the equation, #y=-(x+1)^2+17#, we can see that:

Keep in mind that #h# is negative and not positive even though it appears to be in the equation.

#:.#, the vertex is #(-1,17)#.

Answer 2 · 2015-12-16T05:01:32

Vertex (-1, 17)

y is expressed in vertex form.
Vertex (-1, 17)