# How do you find the vertex and intercepts for #2(x-3)^2=6y+72#?

##### 1 Answer

Feb 8, 2018

#### Explanation:

#"expressing in 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 betwee the vertex and the focus/directrix"#

#rArr2(x-3)^2=6(y+12)#

#rArr(x-3)^2=3(y+12)larrcolor(blue)"in translated form"#

#rArrcolor(magenta)"vertex "=(3,-12)#

#"to obtain the intercepts"#

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

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

#x=0to9=3y+36rArry=-9larrcolor(red)"y-intercept"#

#y=0to(x-3)^2=36#

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

#rArrx-3=+-sqrt36larrcolor(blue)"note plus or minus"#

#rArrx=3+-6#

#rArrx=-3,x=9larrcolor(red)"x-intercepts"#