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for the graph of the quadratic function, identify the coordinates of the vertex: f(x)=-3(x-4)^2+1 Please i need help
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vertex is displayed for you it is \[f(x)=-3(x-4)^2+1\] vertex is \((4,1)\)
so you just take the opposite sign of 4 and then 1 is the y?
The vertex is in the form (x-h) and (y-k). But yes, it is the opposite of the sign in front of what's inside the parenthesis. and the +1 is the y.
\[(x-4)^2\] is always greater than or equal to zero, because it is a square so \(-3(x-4)^2\) is always less than or equal to zero it is only zero if \(x=4\)
therefore the very largest \[-3(x-4)^2\] can be is zero, and if it is zero then \[-3(x-4)^2+1\] is \(1\) that makes the vertex \((4,1)\)
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oh ok thank you both :)
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