How do you find the vertex and intercepts for #x=3(y+3)^2+5#?

1 Answer
Mar 12, 2017

#y_("intercept")=+32#

Vertex#->(x,y)=(5-3)#

As #x_("vertex")=+5# and the graph is of shape #sub# then:
THERE IS NO Y INTERCEPT

Explanation:

Given:#" "x=3(y+3)^2+5#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("What is normally considered as "x" has to be considered as "y)#
Thus INSTEAD OF VERTEX #(x,y)=(-3,5)# WE HAVE VERTEX #(x,y)=(5,-3)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Multiply out the brackets and we have:

#x=3y^2+18ycolor(red)(+32)#

This is a quadratic in y so and as the #y^2# term is positive it is of general shape: #sub#

Thus:

#color(red)(y_("intercept")=+32)#

Vertex#->(x,y)=(5,-3) larr" graph of shape" sub#

Tony B