# Thread: finding points of intersections

1. ## finding points of intersections

Hi, i'm having a bit of a problem here.. I have no idea how to do this.

Show that the curves
$\displaystyle y=2^x$ and $\displaystyle y=1+2x-x^2$ intersect at A(0,1) and B(1,2)

2. Originally Posted by ipokeyou
Hi, i'm having a bit of a problem here.. I have no idea how to do this.

Show that the curves
$\displaystyle y=2^x$ and $\displaystyle y=1+2x-x^2$ intersect at A(0,1) and B(1,2)

for point A, substitute in x = 0 into both equations ... do you get the same y-value?

do the same with point B's x-value.

3. There is probably a better way than what I am about to show you, but I think I know using visualisation geometry.

y=1+2x-x^2 can be factorized as y= -(x-1)(x-1)

If we ignore the negative in this factorization, we know that it is a 'happy face' parabola that it's minimum y value is (1,0).

We also know that because the equation has a +1 (y=1+2x-x^2) it will hit the y axis at (0,1).

However, it is not a happy face parabola because it is a negative (remember we said we would first ignore it?). This therefore makes the whole graph a complete reflection of itself in the same line that it hits the y-axis. Therefore, the line of symmetry is y=1.

Therefore, we know that the highest y value of the graph is (1,2), because the original 'dip' of the graph was 1 below the line of reflection, now it is 1 above it.

So, looking at where they intersect, we can immediately figure out that the first point is (0,1) because whenever you have a power to 0 the value is 1. We can also then easily work out that the point (2,4) is also an intersection because 4 is simply a power of 2.

Kind of long, but I guess the idea is to have a rough drawing of the lines so you can work out the intersections

4. Originally Posted by skeeter
for point A, substitute in x = 0 into both equations ... do you get the same y-value?

do the same with point B's x-value.
yes i did do it that before. but i was wondering how i should solve for x algebraically, like equating them?

5. Originally Posted by Illusion3
There is probably a better way than what I am about to show you, but I think I know using visualisation geometry.

y=1+2x-x^2 can be factorized as y= -(x-1)(x-1)

If we ignore the negative in this factorization, we know that it is a 'happy face' parabola that it's minimum y value is (1,0).

We also know that because the equation has a +1 (y=1+2x-x^2) it will hit the y axis at (0,1).

However, it is not a happy face parabola because it is a negative (remember we said we would first ignore it?). This therefore makes the whole graph a complete reflection of itself in the same line that it hits the y-axis. Therefore, the line of symmetry is y=1.

Therefore, we know that the highest y value of the graph is (1,2), because the original 'dip' of the graph was 1 below the line of reflection, now it is 1 above it.

So, looking at where they intersect, we can immediately figure out that the first point is (0,1) because whenever you have a power to 0 the value is 1. We can also then easily work out that the point (2,4) is also an intersection because 4 is simply a power of 2.

Kind of long, but I guess the idea is to have a rough drawing of the lines so you can work out the intersections
Wow i never thought of it that way.. but yes that makes sense too thanks

6. Originally Posted by Illusion3
y=1+2x-x^2 can be factorized as y= -(x-1)(x-1) ... no, it cannot.
Originally Posted by ipokeyou
yes i did do it that before. but i was wondering how i should solve for x algebraically, like equating them?
$\displaystyle 2^x = 1 + 2x - x^2$

... using elementary algebra, it cannot, except maybe by "observation".

one can use technology (a grapher) to solve for the points of intersection, or one can use advanced methods related to the Lambert W function.