What is the vertex of the parabola defined by the equation y = x2 + 4x + 4?
the formula to find x-coordinate of the vertex is \[\rm x=\frac{-b}{2a}\] a,b,c values are \[\rm Ax^2+Bx+C=0\] where a=leading coefficient b= coefficient of x term c=constant find x and then substitute this x value into the original quadratic equation to find y coordinate
so would it be 2,0?
hint: \(a=1\) and \(b=4\) so: \[{x_V} = \frac{{ - b}}{{2a}} = \frac{{ - 4}}{{2 \cdot 1}} = ...?\]
wherein \(x_V\) stands for x-coordinate of the vertex
1,2?
how did you get that ??
i know there are 4 options: first one is ) 2,0 2nd) 1,2 3rd) idk 4th) idk ;) don't guess! c;
there's 2,0 1,2 -2,0 4,2 0,0
okay please plugin b and a value into the equation just like in 3rd comment on this post and then simplify the fraction
4,2
how did you get that ? what's -4/2 = ?
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