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How do I find the vertex of the equation y= (x+7)^2 - 5 ?

Sagot :

Suziaidn
Follow through with the function (x+7)²
(x+7)² is the same as (x+7)(x+7).
You get x² + 14x + 49
Plug it back into the equation to get 
y=x² + 14x + 49 - 5, which equals y=x² + 14x + 44
The formula for the x point in the vertex is -b/2a. (a is x² and b is x)
So -14/2 = -7 = x
Plug this into the equation.
y=-7² + 14(-7) + 44
y=49 - 98 + 44
y=-5

(-7, -5)

I apologize for the error beforehand--I typed 49 instead of 44 and ended up with 0 instead of -5. 
Konrad509idn
If you have an equation in the vertex form [tex]y=(x-h)^2+k[/tex] the vertex is [tex](h,k)[/tex]

In your case [tex]h=-7,k=-5[/tex] so the vertex is [tex](-7,-5)[/tex]
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