#color(blue)("Method 1")#
Given that the standard form for a quadratic equation is:
#ax^2+bx+c=0#
and: #color(white)(....)x =(-b+-sqrt(b^2-4ac))/(2a)#
Then you could use this to find the #x# intercepts and that #x_("vertex")# is half way between them. That is #color(blue)(-b/(2a))#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Method 2")#
#color(brown)("Use something that is similar to completing the square:")#
#color(green)("When you think about this, it is the same thing as method 1!")#
Write as: #y=2(x^2-14/2x)-5#
Now consider just the brackets
#color(blue)(x_("vertex")=) (-1/2)xx (-14/2)=+14/4 =color(blue)( +3 1/2 )#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Having found #x_("vertex")# we can find the value of #y_("vertex")# by substituting for #x# in the original equation.
#y_("vertex")= 2x^2 -14x-5#
#y_("vertex")= 2(7/2)^2-14(7/2)-5#
#color(blue)(y_("vertex") =)49/2-49-5 = color(blue)(-29 1/2)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#(x_("vertex") , y_("vertex")) -> (3 1/2, -29 1/2)#
