How do you find the vertex and intercepts for #(x − 2)^2 = -12(y − 2)#?

2 Answers
Feb 8, 2018

So the curve has:

Vertex: maximum at #(2,2)#

Y-intercept: #y=13/6#

X-intercepts: #x=2+sqrt6# and #x=2-sqrt6#

Explanation:

#(x-2)^2=-12(y-2)#

Rearrange for y.

#-1/12(x-2)^2=y-2#

#y=-1/12(x-2)^2+2#

From the completed square form, the maximum turning point is at #(2, 2)#

For the y-intercept, let #x=0#

#y=-1/12(-2)^2+2#
#y=13/6#

For the x-intercepts, let #y=0#

#0=-1/12(x-2)^2+2#
#0=(x-2)^2-6#
#6=(x-2)^2#

#x-2=pmsqrt6#

#x=2pmsqrt6#

So the curve has:

Vertex: maximum at #(2,2)#

Y-intercept: #y=13/6#

X-intercepts: #x=2+sqrt6# and #x=2-sqrt6#

graph{(x-2)^2=-12(y-2) [-10, 10, -5, 5]}

Feb 8, 2018

#"see explanation"#

Explanation:

#"the quadratic is in standard translated form"#

#•color(white)(x)(x-h)^2=4p(y-k)#

#"where "(h,k)" are the coordinates of the vertex and p"#

#"is the distance from the vertex to the focus/directrix"#

#rArrcolor(magenta)"vertex "=(2,2)#

#"to find intercepts"#

#• " let x = 0, in the equation for y-intercept"#

#• " let y = 0, in the equation for x-intercepts"#

#x=0to4=-12y+24rArry=5/3larrcolor(red)"y-intercept"#

#y=0to(x-2)^2=24#

#color(blue)"take square root of both sides"#

#x-2=+-sqrt24larrcolor(blue)"note plus or minus"#

#rArrx=2+-2sqrt6larrcolor(red)"x-intercepts"#