How do you solve #y=x^2+5x+4, y=-x-8# using substitution?
1 Answer
There are no points of intersection.
Explanation:
We have the equations;
I'm going to substitute the second equation into the first equation:
and now solve for
I don't see any factors, so I'll shift over to the quadratic formula, with the general formula of:
where we have
In our case we have
It's at this point that we stop - the points of intersection should be Real numbers (
graph{(y-x^2-5x-4)(y+x+8)=0 [-25.66, 25.66, -12.83, 12.83]}