How do you find the VERTEX of a parabola #y=x^2-x-6#?

1 Answer
Meave60
Jul 21, 2015

The vertex is #(1/2-25/4)# or #(1/2, -6 1/4)#.

Explanation:

#y=x^2-x-6# is a quadratic equation in the form of #ax^2+b^2+6#, where #a=1, b=-1, and c=-6#.

The vertex is the minimum or maximum point of the equation. The #x# value can be found using the formula #x=(-b)/(2a)#. The #y# value can be found by substituting the value for #x# into the equation and solving for #y#.

The value of #x#.

#x=(-b)/(2a)=(-(-1))/(2*1)=1/2#

#x=1/2#

The value of #y#.

#y=(1/2)^2-1(1/2)-6# =

#y=1/4-1/2-6# =

The common denominator for the right side is #4#. Multiply each term by a fraction so that it has a denominator of #4#.

#y=1/4-1/2(2/2)-6(4/4)# =

#y=1/4-2/4-24/4# =

#y=-25/4#

Simplify the improper fraction.

#y=-6 1/4#

graph{y=x^2-x-6 [-14.24, 14.23, -7.12, 7.12]}

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