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What is the vertex form of #5y = 13x^2+20x+42#?

Answer 1

Eli Assouline

#y = 13/5(x - -10/13)^2 + 446/65#

Divide both sides by 5:

#y = 13/5x^2 + 4x + 42/5#

The equation is in standard form, #y = ax^2 + bx + c#. In this form the x coordinate, h, of the vertex is:

#h = -b/(2a)#

#h = - 4/(2(13/5)) = -20/26 = -10/13#

The y coordinate, k, of the vertex is the function evaluated at h.

#k = 13/5(-10/13)^2 + 4(-10/13) + 42/5#

#k = 13/5(-10/13)(-10/13) - 40/13 + 42/5#

#k = (-2)(-10/13) - 40/13 + 42/5#

#k = 20/13 - 40/13 + 42/5#

#k = -20/13 + 42/5#

#k = -100/65 + 546/65#

#k = 446/65#

The vertex form of the equation of a parabola is:

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

Substituting in our known values:

#y = 13/5(x - -10/13)^2 + 446/65#

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Answer 2

Kennedy Adams

The vertex form of the quadratic equation 5y = 13x^2 + 20x + 42 is:

y = 13/5 (x + 1) ^ 2 - 13/5

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Answer from HIX Tutor

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