How do you write $y = - 16 x^{2} + 40 x + 4$ $y = -16x^2+40x+4$ into vertex form?

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Answer 1 · 2015-05-07T17:55:01

Vertex form for a parabola is
$y = m {(x - a)}^{2} + b$ $y=m(x-a)^2+b$
with the vertex at $(a, b)$ $(a,b)$

Re-arranging $y = 16 x^{2} + 40 x + 4$ $y=16x^2+40x+4$ into vertex form:

$y = - 16 (x^{2} - \frac{5}{2} x) + 4 extract the m factor$ $y=-16(x^2-5/2x) +4 " extract the "m" factor"$

$y = - 16 (x^{2} - \frac{5}{2} x + {(\frac{5}{4})}^{2}) + 25 + 4 complete the square$ $y=-16(x^2-5/2x+(5/4)^2) +25 +4 " complete the square"$

$y = - 16 {(x - \frac{5}{4})}^{2} + 29$ $y = -16(x-5/4)^2+29$

How do you write y = -16x^2+40x+4y=−16x2+40x+4 y = -16x^2+40x+4 into vertex form?