What is the vertex of #x – 4y^2 + 16y – 19 = 0#?

1 Answer
Tony B
Oct 22, 2017

Vertex #->(x,y)=(3,2)#

Explanation:

This is a quadratic in #y# instead of #x#. Thus the graph is like one done in #x# but rotated #90^0#. So you would have an axis of symmetry that is parallel to the y-axis.

As the coefficient of #y^2# is positive the graph is of form #sub#

Write as #x=4y^2-16y+19" "............. Equation(1)#

Now write as #x=4(y^2-16/4x)+19" ".................Equation(1_a)#

Normally I would use the cheat of
#x_("vertex")=(-1/2)xxb/a #

However in this case we have
#y_("vertex")=(-1/2)xx(-16/4) = +2#

Note that this is part of the process of completing the square.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So by substitution for #y#

#x_("vertex")=4(+2)^2-16(+2)+19 = 3#

Vertex #->(x,y)=(3,2)#
Tony B

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