At what points do the functions y=x2−x and y=sinπx intersect?

1 Answer
George C.
Jan 24, 2017

These two equations intersect at the points (0,0) and (1,0)

Explanation:

First, let us take a look at:

y=x2−x

We can factor this as:

y=x(x−1)

so this quadratic has x intercepts at (0,0) and (1,0)

It has minimum value at the midpoint of these two x coordinates, where x=12:

y=12(12−1)=−14

So note that y=x2−x<0 for all x∈(0,1)

The intersections of y=x2−x with the horizontal line y=1 are at the points given by solving:

0=x2−x−1

0=x2−x+14−54

0=(x−12)2−(√52)2

0=(x−12−√52)(x−12+√52)

That is:

x=12±√52

Note that 12+√52≈1.618∈(1,2)

Note that 12−√52≈−0.618∈(−1,0)

So x2−x∈(0,1] when x∈[12−√52,0) or x∈(1,12+√52]

Outside these intervals, x2−x>1 so cannot be equal to sin(πx)

Now consider y=sin(πx) in each of these intervals:

  • If x∈[12−√2,0) then sin(πx)<0

  • If x=0 then sin(πx)=0=x2−x

  • If x∈(0,1) then sin(πx)>0

  • If x=1 then sin(πx)=0=x2−x

  • If x∈(1,12+√52) then sin(πx)<0

So in each of the intervals [12−√52,0), (0,1) and (1,12+√52] the two functions have opposite signs and cannot intersect.

So the only two points of intersection are:

(x,y)=(0,0)

(x,y)=(1,0)

graph{(y-x^2+x)(y - sin(pix)) = 0 [-2.105, 2.895, -1.19, 1.31]}

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