What is the vertex and the equation of the axis of symmetry graph of #y=x^2-6x-7#?

Question

Kennedy Adams · Accepted Answer

The vertex is at #(3, -16)# and the axis of symmetry is #x=3#.

First, the EASY WAY to do this problem. For ANY quadratic equation in standard form #y=ax^2+bx+c# the vertex is located at #(-b/(2a),c-b^2/(4a))#. In this case #a=1#, #b=-6#, and #c=-7#, so the vertex is at #(-(-6)/(2*1),-7-(-6)^2/(4*1))=(3, -16)#. But suppose you didn't know these formulae. Then the easiest way to get the vertex information is to convert the standard form quadratic expression into the vertex form #y=a(x-k)^2+h# by completing the square. The vertex will be at #(k, h)#. #y=x^2-6x-7=x^2-6x+9-16=(x-3)^2-16#. Again we see that the vertex is at #(3,-16)#. The axis of symmetry for a parabola is always the vertical line containing the vertex (#x=k#), or in this case #x=3#. graph{x^2-6x-7 [-10, 10, -20, 5]}

Isaiah Anthony · Answer

A different approach:

Axis of symmetry #->x=3#

Vertex #->(x,y)=(3,-16)#

Given: #y=x^2color(red)(-6)x-7#

What I am about to do is part of the process of completing the square.

#y=a(x+color(red)(b)/(2a))^2+k+c#

In this case #a=+1# so we ignore it.

Note that #color(red)(b=-6)#

#x_("vertex") = x_("axis of symmetry")=(-1/2)xxcolor(red)(b)#

# color(white)("dddddddddddddddddddd") (-1/2)color(red)(xx(-6))=+3#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Substitute for #x=+3#

#y=x^2-6x-7color(white)("dddd")->color(white)("dddd")y=3^2-6(3)-7#

#color(white)("d"dddddddddddddddd.)->color(white)("dddd")y=-16#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Axis of symmetry #->x=3#

Vertex #->(x,y)=(3,-16)#