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What is the vertex form of #5y = -3x^2-2x+2#?

Answer 1

Eli Assouline

#y = -3/5(x + 1/3)^2 + 7/15#

Given: #5y = -3x^2 - 2x + 2#

Vertex form: #y = a(x-h)^2 + k#,

where the vertex is #(h, k)# and #a# is a constant.

Complete the square to find the vertex form.

First divide by #5# to make #y = #:

#y = (-3/5x^2 - 2/5x )+ 2/5#

Factor out #-3/5# so we only have an #x^2#:

#y = -3/5(x^2 + 2/3 x) + 2/5#

To complete the square we need to multiply #1/2 * 2/3 = 1/3#. We also need to subtract the squared term that we added:

#-3/5 (x+1/3)(x+1/3) = color(red)(-3/5)(x^2 + 2/3x " "color(red)( + 1/9))#

#y = -3/5(x + 1/3)^2 + 2/5 - (color(red)(-3/5 (1/3)^2))#

#y = -3/5(x + 1/3)^2 + 3/3*2/5 + 1/15#

#y = -3/5(x + 1/3)^2 + 7/15#

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Answer 2

Julia Adams

The vertex form of the given equation is: [ y = a(x - h)^2 + k ] where: [ a = -3 ] [ h = -\frac{b}{2a} = -\frac{-2}{2(-3)} = \frac{1}{3} ] [ k = 2 ]

Therefore, the vertex form is: [ y = -3(x - \frac{1}{3})^2 + 2 ]

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Answer from HIX Tutor

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