How do you find the vertex of the parabola #y=2x^2-8x+7#?

2 Answers
Feb 9, 2015

You can find the coordinates of your vertex either using a mnemonic or the derivative:

1) Mnemonic: the coordinates of the vertex of a parabola in the form:
#y=ax^2+bx+c# are:
#x_v=-b/(2a)#
#y_v=-Delta/(4a)# where #Delta=b^2-4ac#
in your case:
#x_v=2#
#y_v=-1#

2) Derivative: determine the derivative of your function and set it equal to zero:
#y'=4x-8#
#4x-8=0# gived #x=x_v=2#
substituting back in your function:
#y=y_v=2*4-8*2+7=-1#

Graphically:
graph{2x^2-8x+7 [-4.385, 4.386, -2.19, 2.193]}

Mar 5, 2016

A slight variant on method

#color(blue)("Vertex" ->(x,y)->(2,-1)#

Explanation:

Given:#" "2x^2-8x+7#

Write as:#" " 2(x^2-color(red)(8/2)x)+7#

Consider the #color(red)(-8/2)" from "-8/2x#

#color(blue)(x_("vertex")=(-1/2)xx(color(red)(-8/2)) = +8/4 = 2)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Substitute #x=color(magenta)(2)# in the original equation to find #y_("vertex")#

#y=2x^2-8x+7" "->" "y_("vertex")=2(color(magenta)(2))^2-8(color(magenta)(2))+7#

#color(blue)(y_("vertex")= 8-16+7=-1)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Vertex" ->(x,y)->(2,-1)#

Tony B