What is the vertex form of #y= x^2/4 - x - 4 #?

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Answer 1 · 2017-08-27T15:04:33

#y = 1/4(x-2)^2-5#

The given equation

#y= x^2/4 - x - 4" [1]"#

is in standard form:

#y = ax^2+ bx + c#

where #a = 1/4, b = -1 and c = -4#

Here is a graph of the given equation:

graph{x^2/4 - x - 4 [-8.55, 11.45, -6.72, 3.28]}

The vertex form for a parabola of this type is:

#y = a(x-h)^2+k" [2]"#

where #(h,k)# is the vertex.

We know that "a" in the standard form is the same as the vertex form, therefore, we substitute #1/4# for "a" into equation [2]:

#y = 1/4(x-h)^2+k" [3]"#

To find the value of #h#, we use the formula:

#h = -b/(2a)#

Substituting in the values for "a" and "b":

#h = - (-1)/(2(1/4))#

#h = 2#

Substitute 2 for #h# into equation [3]:

#y = 1/4(x-2)^2+k" [4]"#

To find the value of k, we evaluate the given equation at #x = h = 2#:

#k= (2)^2/4 - 2 - 4#

#k = 1 - 2 - 4#

#k = -5#

Substitute -5 for #k# into equation [4]:

#y = 1/4(x-2)^2-5#

Here is a graph of the vertex form:

graph{1/4(x-2)^2-5 [-8.55, 11.45, -6.72, 3.28]}

Please observe that the two graphs are identical.