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Which is a quadratic equation with a vertex of (4,-3_ and passes through point (6,2)? The answer is y= 5/4 (x-4^2 -3 .. I know that h=4 ad k= -3 so idk how to get the 5/4, so i need to show the work
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\[f(x)=a(x-4)-3\] solve for \(a\) by solving \[f(6)=2=a(6-4)-3\]
you get \[2=a(6-4)^2-3\]
whoa typos on the first two
\[f(x)=a(x-4)^2-3\\ f(6)=a(6-4)^2-3=2\]
\[4a-3=2\] solve for \(a\)
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Solve for a? how i do that?
How to you get 4a-3=2?
@satellite73
Yeah I got it but how did you get 4a-3=2?
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