How do you write #y=3x^2 + x - 2# in vertex form?

1 Answer
Aug 6, 2017

Use the form:

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

And the formulas:

#a =a" [2]"#

#h=-b/(2a)" [3]"#

#k = y(h)" [4]"#

Explanation:

The given equation:

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

is in the standard form:

#y = ax^2+bx+c#

where #a = 3#, #b=1#, and #c = -2#

We shall use formula [2] to substitute 3 for "a" into equation [1]:

#y=3(x-h)^2+k" [1.1]"#

Use formula [3] to compute the value of h:

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

#h = -1/6#

Substitute #-1/6# for h into equation [1.1]:

#y=3(x-(-1/6))^2+k" [1.2]"#

Note: You can write the above as #x+1/6# but doing so can cause an error, when you are asked for the x coordinate of the vertex, therefore, I do not recommend it.

Use formula [4] to determine the value of k:

#k = y(h)#

#k = y(-1/6)#

#k = 3(-1/6)^2 + (-1/6) - 2#

#k = -25/12#

Substitute #-25/12# for k into equation [1.2]:

#y=3(x-(-1/6))^2-25/12" [1.3]"#

Equation [1.3] is the answer.