Question

for the graph of the quadratic function, identify the coordinates of the vertex: f(x)=-3(x-4)^2+1 Please i need help

Answer 1

vertex is displayed for you
it is
\[f(x)=-3(x-4)^2+1\] vertex is \((4,1)\)

Answer 2

so you just take the opposite sign of 4 and then 1 is the y?

Answer 3

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.

Answer 4

\[(x-4)^2\] 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 \(x=4\)

Answer 5

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 \((4,1)\)

Answer 6

oh ok thank you both :)