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Precalculus
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"Find the vertex of the graph of the function" f(x)=(x+6)^2-5
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Can you simplify it into f(x)=Ax+By+C form?
*Ax^2+Bx+C, I was thinking of standard form for some reason
how would that be finding the vertex
?
the equation is already in the vertex form for parabolas therefore the vertex of your equation of f(x)=(x+6)^2-5 is (-6, 5) because when f(x)=a(x-h)^2-k, (h, k) is the vertex hope this helped :)
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