What is the vertex form of $y=4x^2+4x+1$ ?

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Answer 1 · 2017-10-11T12:22:22

Vertex form of equation is $y = 4(x+ 0.5)^2+0$

$y = 4x^2 +4x+1 or y = 4(x^2 +x) +1$

$y = 4(x^2 +x + 0.5^2) -1+1; [4*0.5^2=1]$ or

$y = 4(x+ 0.5)^2+0$ . Comparing with vertex form of equation

$y = a(x- h)^2+k ; (h,k)$ being vertex, we find

$h=-0.5 and k= 0$ . So vertex is at $(-0.5,0)$ and

vertex form of equation is $y = 4(x+ 0.5)^2+0$ [Ans]

What is the vertex form of y=4x^2+4x+1 y=4x^2+4x+1 ?