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Which of the following is the vertex of the quadratic equation y = -4(x - 6)^2 + 2?
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for \(y=a(x-b)^2+c\) we have that \((b,c)\) is the vertex point.
how do you find c
-6,2
\(y=a(x-b)^2+c \iff y=-4(x-6)^2+2\)
why -6?
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notice the form has (x-b) so since you have (x-6) we have b=6
i dont know
if you had (x+6) then b=-6
we want it to be \((\small x\large-\small b)\)
so the answer is (6,2)
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ohh
if you had \(-4(x+6)^2+2\) then the answer would be \((-6,2)\)
because for our form to work we need it to be a minus sign, so its \(-4(x-(-6))^2+2\)
do a few more like this and you will get what I'm saying.
thanks
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