What is the vertex of the parabola #y=(x+8)^2+1#?

1 Answer
Mar 8, 2016

#color(blue)(x_("vertex")=-8)#
I have taken you up to appoint where you should be able to finish it off.

Explanation:

Standard form #y=ax^2+bx+c#

Write as:#" "y=a(x^2 +b/ax) +c#

Then #x_("vertex")=(-1/2)xxb/a#

Expanding the brackets

#y=x^2+16x +84+1#

In your case #a=1" so "b/a =16/1#

Apply # (-1/2)xx16= -8#

#color(blue)(x_("vertex")=-8)#

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Find #y_("vertex")" " #by substitution

#color(brown)(y=x^2+16x +85)color(green)( -> y=(-8)^2+16(-8) +85)#

I will let you finish this bit

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Tony B