What is the vertex form of #y= 6x^2 - 4x - 24 #?

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Answer 1 · 2016-07-02T17:08:24

#y = 6(x-1/3)^2 - 24 2/3#

The vertex is at #(1/3 . -24 2/3)#

If you write a quadratic in the form

#a(x +b)^2 +c#, then the vertex is #(-b,c)#

Use the process of completing the square to get this form:

#y = 6x^2 - 4x -24#

Factor out the 6 to make #6x^2# into #"x^2#

#y = 6(x^2 -(2x)/3 - 4)" " 4/6 = 2/3#

Find half of #2/3# .................................#2/3 ÷ 2 = 1/3#

square it....... #(1/3)^2# and add it and subtract it.

#y = 6[x^2 -(2x)/3 color(red)(+ (1/3)^2) - 4 color(red)(- (1/3)^2)]#

Write the first 3 terms as the square of a binomial

#y = 6[(x-1/3)^2 - 4 1/9]#

Multiply the 6 into the bracket to get the vertex form.

#y = 6(x-1/3)^2 - 24 2/3#

The vertex is at #(1/3 . -24 2/3)#