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

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Answer 1 · 2016-03-08T18:44:13

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

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

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Tony B

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