What is the vertex form of $y=5x^2+17x+13$ ?

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Answer 1 · 2017-11-22T04:28:28

When given an equation of the form:

$y=ax^2+bx+c$

The vertex form is:

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

where $h = -b/(2a) and k = ah^2+bh+c$

Given:

$y=5x^2+17x+13$

Please observe that $a = 5, b = 17, and c = 13$

Substitute 5 for a in the vertex form:

$y = 5(x-h)^2+k$

Compute the value of h:

$h = -17/(2(5))$

$h = -17/10$

Substitute the value of h into the vertex form:

$y = 5(x-(-17/10))^2+k$

Compute the value of k:

$k = 5(-17/10)^2+17(-17/10)+13$

$k = -29/20$

Substitute the value of k into the vertex form:

$y = 5(x-(-17/10))^2+ (-29/20)$

The above equation is the vertex form.

What is the vertex form of y=5x^2+17x+13y=5x^2+17x+13 ?