What is the the vertex of $y = - 4 x^{2} + 2 x + 1$ $y =-4x^2+2x+1$ ?

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Answer 1 · 2016-01-04T03:28:05

$(\frac{1}{4}, \frac{5}{4})$ $(1/4, 5/4)$

The vertex form of a quadratic equation is
$y = a {(x - h)}^{2} + k$ $y = a(x-h)^2 +k$

where $(h, k)$ $(h, k)$ is the vertex of the quadratic.

To put the equation into vertex form, we can use a process called completing the square.

$y = - 4 x^{2} + 2 x + 1$ $y = -4x^2 + 2x + 1$

$= - 4 (x^{2} - \frac{1}{2} x) + 1$ $= -4(x^2 - 1/2x) + 1$

$= - 4 (x^{2} - \frac{1}{2} x + \frac{1}{16} - \frac{1}{16}) + 1$ $= -4(x^2 -1/2x +1/16 - 1/16) +1$

$= - 4 (x^{2} - \frac{1}{2} x + \frac{1}{16}) + \frac{1}{4} + 1$ $=-4(x^2 - 1/2x +1/16) + 1/4 + 1$

$= 4 {(x - \frac{1}{4})}^{2} + \frac{5}{4}$ $=4(x-1/4)^2 + 5/4$

Thus the vertex is $(\frac{1}{4}, \frac{5}{4})$ $(1/4, 5/4)$

What is the the vertex of y=−4x2+2x+1y =-4x^2+2x+1 ?