How do you find the axis of symmetry and vertex point of the function: #y=4x^2-4x-15#?

1 Answer
Alan P.
Oct 9, 2015

Vertex: #(1/2,-16)#
Axis of symmetry: #x=1/2#

Explanation:

Given
#color(white)("XXX")y=4x^2-4x-15#

Part 1: The Vertex
Convert into vertex form (#g(x)=m(x-a)^2+b# with vertex at #(a,b)#)

#color(white)("XXX")#Extract the #m#
#color(white)("XXX")y)=4(x^2-1x) - 15#

#color(white)("XXX")#Complete the square
#color(white)("XXX")y= 4(x^2-1x+(1/2)^2)-15 - 4(1/2)^2)#

#color(white)("XXX")#Re-write as a squared binomial of form #(x-a)^2# and simply
#color(white)("XXX")y=4(x-(1/2))^2+(-16)#

This is in vertex form with the vertex at #(1/2,-16)#

Part 2: The Axis of Symmetry
An parabolic equation in the form:
#color(white)("XXX")y=ax^2+bx+c#
has a vertical axis (i.e. #x=c# for some constant #c#) through the vertex.

Since the given #y=)4x^2+4x-15# is of this form and the vertex is at #(x,y)=(1/2,-16)#
the axis of symmetry is
#color(white)("XXX")x=(1/2)#
graph{4x^2-4x-15 [-5.45, 14.56, -18.64, -8.64]}

Related questions
See all questions in Vertex Form of a Quadratic Equation
Impact of this question
1146 views around the world
You can reuse this answer
Creative Commons License