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

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Answer 1 · 2015-12-25T11:24:21

There are many ways of finding the vertex form of this type quadratic functions. An easy method is given below.

If we have #y =ax^2+bx+c# and to write it in vertex form we do the following steps.

If the vertex is #(h,k)# then #h=(-b/(2a))# and #k=a(h)^2+b(h)+c#

The vertex form is y=a(x-h)^2 + k.

Now let us use the same with our question.

#y=-x^2-3x+5#

Comparing it with # y=ax^2+bx+c # we get #a=-1#, #b=-3#, #c=5#

#h=-b/(2a)#
#h=-(-3)/(2(-1))#
#h=-3/2#

#k=-(-3/2)^2-3(-3/2)+5#
#k=-9/4 +9/2 + 5#
#k=+9/4 + 5#
#k= 9/4 + 20/4#
#k=29/4#

#y=-(x-(-3/2))^2 + 29/4#

#y=-(x+3/2)^ 2+ 29/4# is the vertex form