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

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Answer 1 · 2016-10-16T20:02:24

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

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:

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

What is the vertex form of 5y = 13x^2+20x+425y = 13x^2+20x+42?