two functions points of intersection?

Question

If you are given two functions how do you find their points of intersection?

The functions I'm personally working with are y=xsquared and y=4x-(x squared). I have the answer: (0,0) and (2, 4), I just don't know how work the problem.

Help would be greatly appreciated!

Answer 1

You can think of it this problem as where do these 2 functions meet/intersect or where are their values equal... so...

y=4x-x^2 <=> y=x^2

so... set both equations equal to one another

4x-x^2 = x^2

-2x^2 + 4x = 0
-2x(x^2 - 4) = 0

-2x = 0 OR x^2 - 4 = 0

x = 0 or x^2 = 4 , x = +/- 2


Now that you have the values for x, you can solve for y by substituting 0, -2,2 in either y = x^2 or y = 4x - x^2

Good luck, hope this helps...

Answer 2

sqrt(4x) = 3x sq. the two factors (sqrt(4x))^2 = (3x)^2 4x = 9x^2 9x^2 - 4x = 0 element x(9x - 4) = 0 x = 0 or 9x - 4 = 0 x = 0 or x = 4/9 ~~~~~~~~~~~~~~~~~~~~~~~~ examine answer: First examine x = 0: sqrt(4(0)) = 3(0) 0 = 0 (authentic) Now, examine x = 4/9 sqrt((4)*(4/9)) = 3(4/9) sqrt(sixteen/9) = 12/9 4/3 = 4/3 (authentic) hence, the answer is x = 0 and x = 4/9 So, then the factors of intersection are (0,0) and (4/9,4/3)

Answer 3

solve them to get the points of intersection
y=x^2=4x-x^2
2x^2-4x=0
2x(x-2)=0
x=0 or 2
when x=0 y=0
when x=2 y=4
you are right

Answer 4

if they intercect, they are equal ..... set them equal

y = x^2 and y = 4x-x^2 so ( y=y ) ...so ( x^2 = 4x-x^2 )
bring all to one side (quadradic) 2(x^2) - 4x = 0

factor out (2x) .... (2x) * (x-2) = zero 2x=0 x=0 .... x-2 = 0 x=2

x=(0,2) when x=0,y=0 (0,0) when x=2,y=4 (2,4)

your solution is (0,0) & (2,4)