Thread: Find the x and y intercepts

3x^2+12x+11
I can get it down to 3(x+2)^2-1 by completing the square. But I can't remember how to find the x intercepts algebraically. same with the y intercept I somehow have to get -2+or- sqrt 3/3? Where the heck does the sqrt 3 come from? And I cannot use a calc or quadratic formula.

Shuldn't it be y = 3x^2+12x+11
well won't +11 be the y-intercept (y-intercept is when x=0, 0 + 0 + 11 = 11, y-intercept therfore = 11!!)
and as you simplified it to
3(x+2)^2-1
the x-intercept is attained when y = 0 , so when y = 0
0 = 3(x+2)^2-1
1 = 3(x+2)^2
1/3 = (x+2)^2
+-sqrt (1/3) = x + 2 (square root of a number can have a positive value or a negative value)
x = +-sqrt (1/3) - 2

PS +- = plus or minus, when written they usualy go under one another, but the computer can't do that for me