What is the vertex form of #y=5x^2 + 4x - 3?

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Answer 1 · 2015-11-16T00:36:44

$y=5(x+2/5)^2-19/5$

A parabola has the form:
$y=ax^2+bx+c$
$y=5x^2+4x-3$

The vertex form of a parabola is:
$y=a(x-h)^2+k$

To find the vertex $(h,k)$ of a parabola we use:

$h =-b/{2a}=-4/{2*5}=-4/10=-2/5=-0.4$

And to find the y coordinate we just evaluate in the original function with the value we just found:

$k=f(-2/5)=5(-2/5)^2+4(-2/5)-3=-3.8$

The vertex is $(-0.4,-3.8)$ . Rewriting the equation using the vertex form, we have:

$y=5(x-(-0.4))^2+(-3.8)$
$y=5(x+0.4)^2-3.8$

If you want to have it in a nicer way:

$y=5(x+2/5)^2-19/5$

graph{5(x+2/5)^2-19/5 [-10, 10, -5, 5]}
Click on the graph to see its vertex