What is the vertex form of $2y=5x^2+8x − 4.$ ?

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Answer 1 · 2017-06-25T09:27:06

The vertex form is $y=5/2(x+4/5)^2-18/5$

Let simplify the equation by completing the squares

$2y=5x^2+8x-4$

Dividing by $2$

$y=5/2x^2+4x-2$

$=5/2(x^2+8/5x)-2$

Completing the squares, adding half of the coefficient of $x$ to the square and removing it

$y=5/2(x^2+8/5x+4^2/5^2)-2-5/2*4^2/5^2$

$y=5/2(x^2+8/5x+16/25)-2-8/5$

Factorising

$y=5/2(x+4/5)^2-18/5$

This is the vertex form

graph{y=5/2(x+4/5)^2-18/5 [-8.89, 8.89, -4.444, 4.445]}

What is the vertex form of 2y=5x^2+8x − 4.2y=5x^2+8x − 4.?