What is the coordinates of the point of intersection of the curve #y=1/(x+2) +3# and the line #y=-x#?

1 Answer
Jan 18, 2018

Those two curves do not intersect.

Explanation:

To find an intersection we try to find an #x#-value where both equations give us the same #y# value.

So we want to solve:

#1/(x+2) + 3 = -x#

We'll begin by clearing the fraction. Multiply both sides of the equation by #(x+2)#

#(x+2)(1/(x+2) + 3) = -x(x+2)#

#1 + 3(x+2) = -x(x+2)#

#1+3x+6 = -x^2-2x#

We see an #x^2#, so this is a qudratic equation. The usual way of writing a quadratic equation is to have all of the terms on one side equal to #0# on the other, so let's make it like that.

#x^2+5x+7 = 0#.

We try to solve by factoring, but we don't see how to factor this, so either solve by completing the square or solve using the quadratic formula.

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#x = (-(5)+-sqrt((5)^2-4(1)(7)))/(2(1))#

But now we notice (if we didn't notice before) that

#sqrt(b^2-4ac) = sqrt((5)^2-4(1)(7)) = sqrt(25-28) = sqrt(-3)#.

So there is no solution in the real numbers.

Therefore, there is no point of intersection for the curves.

Here is the graph. Of course, #y=-x# is the straight line.

graph{(y+x)(y-1/(x+2)-3)=0 [-17.02, 15.01, -5.2, 10.82]}