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vfiekz [6]
3 years ago
7

Which equation is y = –6x2 + 3x + 2 rewritten in vertex form?

Mathematics
2 answers:
slava [35]3 years ago
8 0

we have

y = -6x^{2} + 3x + 2

we know that

the vertex form of the vertical parabola equation is equal to

y=a(x-h)^{2} +k

where

(h,k) is the vertex of the parabola

To find the equation rewritten in vertex form let's factor the equation

<u>Factor the leading coefficient </u>

y = -6(x^{2} - 0.5x) + 2

<u>Complete the square. Remember to balance the equation by adding the same constants to each side</u>

y = -6(x^{2} - 0.5x+0.0625-0.0625) + 2

y = -6(x^{2} - 0.5x+0.0625) + 2 +0.375

y = -6(x^{2} - 0.5x+0.0625) + 2.375

<u>Rewrite as perfect squares</u>

y = -6(x-0.25)^{2} + 2.375

therefore

the answer is

y = -6(x-0.25)^{2} + 2.375

Annette [7]3 years ago
4 0
The answer is y = -6( x- \frac{1}{4}) ^{2}+3

Regular form: y = ax² + bx + c
Vertex form: y = a(x - h)² + k
(h, k) - vertex

y = -6 x^{2} +3x+2 \\ y -2=-6 x^{2} +3x \\  y-2+6*=-6* x^{2} +6* \frac{1}{2} x \\  \\ y-2-6* \frac{1}{16} =-6* x^{2} +6* \frac{1}{2} x -6* \frac{1}{16} \\  \\ &#10;y -2 -\frac{6}{16} =-6( x^{2} -\frac{1}{2} x+\frac{1}{16}) \\  \\ &#10;y- \frac{2*16}{16} -\frac{6}{16} =-6( x^{2} -\frac{1}{2} x+(\frac{1}{4})^{2} ) \\  \\ &#10;y -  \frac{32}{16} -\frac{6}{16}  = -6( x- \frac{1}{4}) ^{2}  \\  \\ &#10;y -  \frac{32+6}{16} = -6( x- \frac{1}{4}) ^{2} \\  \\ &#10;y -  \frac{48}{16} =  -6( x- \frac{1}{4}) ^{2} \\  \\

y -  3=  -6( x- \frac{1}{4}) ^{2} \\  \\ &#10;y  =  -6( x- \frac{1}{4}) ^{2}+3&#10;

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