What is the vertex form of $2y=10x^2+7x-3$ ?

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Answer 1 · 2018-05-05T13:08:19

$color(blue)(y=5(x+7/20)^2-169/80)$

$2y=10x^2+7x-3$

Divide by 2:

$y=5x^2+7/2x-3/2$

We now have the form:

$color(red)(y=ax^2+bx+c)$

We need the form:

$color(red)(y=a(x-h)^2+k)$

Where:

$bba color(white)(8888)$ is the coefficient of $x^2$

$bbh color(white)(8888)$ is the axis of symmetry.

$bbk color(white)(8888)$ is the maximum or minimum value of the function.

It can be shown that:

$h=-b/(2a)color(white)(8888)$ and $color(white)(8888)k=f(h)$

$:.$

$h=-(7/2)/(2(5))=-7/20$

$k=f(h)=5(-7/20)^2+7/2(-7/20)-3/2$

$color(white)(8888)=245/400-49/40-3/2$

$color(white)(8888)= 49/80-49/40-3/2$

$color(white)(8888) =(49-98-120)/80=-169/80$

Vertex form:

$y=5(x+7/20)^2-169/80$

What is the vertex form of 2y=10x^2+7x-32y=10x^2+7x-3?