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.