What are the points of intersection of the graphs of the two functions?

f(x)=g(x)

f(x)=x^2+6x+1
g(x)=1

I set them equal and then got x^2+6x=0, but where do I go from there? Help please. = [

Question

What are the points of intersection of the graphs of the two functions?

f(x)=g(x)

f(x)=x^2+6x+1
g(x)=1

I set them equal and then got x^2+6x=0, but where do I go from there? Help please. = [

unklerhaukus · Answer

good start$$x^2+6x=0$$$$x^2 =-6x$$$$x=-6,~~~~-x=-6$$$$~~~~~~~~~~~~~~~~~~~~~~~x=6$$
$$x=±6 $$

anonymous · Answer

oh!!! thanks so much!! = ]

unklerhaukus · Answer

actally i think i have made a mistake

unklerhaukus · Answer

i cant see where it is though

anonymous · Answer

x=-6 and x=0.
x^2+6x=0
x(x+6)=0
x=0 and x=-6

unklerhaukus · Answer

shankvee has the right answer

anonymous · Answer

oh i see... thanks shankvee

unklerhaukus · Answer

i still cant see what my mistake was

anonymous · Answer

I think it was because it was a x^2 instead of x just being a variable to a constant...maybe? idk. Wouldn't the left side of the equation would of had to of been something like (2x+4)^2 in order for the solving using the square root?

unklerhaukus · Answer

i wasnt taking square root thoug, i was dividing by x

anonymous · Answer

x^2+6x =0
x( x + 6) = 0
x = 0, x = -6

anonymous · Answer

so when it asks for the intersection, what do i plug in for g(x)=1 since the x-intercepts are 0,-6? just multiply?

anonymous · Answer

nevermind, the y intercept would just be 1

anonymous · Answer

It intersects when x=0 and x=-6. If you want the points they are (0,1) and (-6,1).

anonymous · Answer

thanks!