3) y=(x+1)² +6
well, set the functions equal to each other. so for the first one
Then rearange so that you get a quadratic on one side and zero on the other and solve for x like a normal quadratic equation. Then substitue your values for x into either of the original expressions to get the corresponding values of y.
well you got x=0. that is the value of x which the curves intersect, at a point. i.e, both curves are at the same point an x=0. So substitue x=0 into either equation to get 100. Then the coordinates of the intersection are (0,100). In that case there is one intersection, but there may well be two points. (or none, but thats a bit silly)