Question 471502
find the coordinates of the points of intersection of the parabola  y=x2 and the line y=x+2
<pre>
Here is the graph of the parabola y=x²

{{{drawing(400,300,-6,6,-3,6,

graph(400,300,-6,6,-3,6,x^2) )}}}

Here is the graph of the line y=x+2:

{{{drawing(400,300,-6,6,-3,6,

graph(400,300,-6,6,-3,6,x^2), line(-7,-5,7,9)


 )}}}

Now we'll find the coordinates of those two points where
they intersect:

{{{system(y=x^2,y=x+2)}}}

Substitute {{{x^2}}} for y in the second equation:

{{{x^2=x+2}}}

Get 0 on the right:

{{{x^2-x-2=0}}}

Factor the left side:

{{{(x-2)(x+1)=0}}}

Use the zero factor principle:

x-2 = 0     x+1 = 0
  x = 2       x = -1

Now we have to find the y coordinates to go
with each of these x-coordinates:


Find the y-coordinate that goes with x = 2
by substituting in

y = x+2
y = 2+2
y = 4

So one point is (x,y) = (2,4)


------------

Find the y-coordinate that goes with x = -1
by substituting in

y = x+2
y = -1+2
y = 1

So the other point is (x,y) = (-1,1)

And if we look at the graph we see that it checks
with those:

{{{drawing(400,300,-6,6,-3,6,

graph(400,300,-6,6,-3,6,x^2), line(-7,-5,7,9),
locate(-2.5,1.2,"(-1,1)"), locate(2,4,"(2,4)")

 )}}}

Edwin</pre>