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"#