Intersection of a Parabola and a Line

Date: 10/19/2003 at 02:53:23
From: Joey
Subject: polynomial

What equation would you use to determine the x-coordinates of the
points of intersection of a quadractic function ax^2 + bx + c with 
the line y = x?  Explain graphically why this procedure is equivalent
to getting the roots of a quadratic equation.

Date: 10/19/2003 at 03:48:08
From: Doctor Luis
Subject: Re: polynomial

Hi Joey,

The intersection between the two curves

  y = x
  y = ax^2 + bx + c

is obtained by setting them equal to each other since each equals y: 

  ax^2 + bx + c = x

Then, you can subtract x from both sides, giving you

  ax^2 + (b-1)x + c = 0

which is a quadratic equation.  You can see that this quadratic will
give you 2, 1, or 0 real solutions, according to whether the
discriminant (b-1)^2 - 4ac is positive, zero, or negative.

Geometrically speaking, the number of solutions represents the number
of times the line y = x intersects the parabola y = ax^2 + bx + c. 

In the following graph, I've plotted the line y = x, along with three
other parabolas representing the three possible cases of two, one, or
zero intersections:

   

I hope this helped!  Let us know if you have any more questions.

- Doctor Luis, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 10/19/2003 at 04:06:40
From: Joey
Subject: polynomial

Thanks for the quick response.  I can see how finding the points of
intersection involves solving a quadratic equation.  But what if the
line is y = k instead of y = x?  Does that make a difference?

Date: 10/19/2003 at 04:52:12
From: Doctor Luis
Subject: Re: polynomial

Hi Joey,

Yes, it does make a difference.

For a line y = k, where k is a constant, we still have three 
possibilities for the number of intersections (two intersections, one 
intersection, zero intersections), but the quadratic equation you have 
to solve is different.

  y = k
  y = ax^2 + bx + c

The intersections satisfy

  ax^2 + bx + c = k

  ax^2 + bx + (c-k) = 0

Therefore, you will have 2, 1, or 0 intersections according to whether 
the discriminant b^2 - 4a(c-k) in the quadratic equation above is
positive, zero, or negative.

In the following graph, I've plotted a line y = k along with three
other parabolas representing these possibilities:

   

Let us know if you have any more questions.

- Doctor Luis, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 10/19/2003 at 16:41:54
From: Joey
Subject: polynomial

Thank you for answering my questions.  I have one more.  Could you
explain why the quadratic formula calculates the x-coordinates of the 
points of intersection of a second degree polynomial with the x-axis?

Date: 10/19/2003 at 18:11:29
From: Doctor Luis
Subject: Re: polynomial

Hi Joey,

The x-axis is just the line y = 0.  See my previous email with k = 0.
The horizontal line is now the x-axis, and again the parabola can 
intersect it in 2, 1 or 0 places.  The equations are:

  y = 0
  y = ax^2 + bx + c

Setting them equal, ax^2 + bx + c = 0 and this is the "normal" 
quadratic equation you solve using the quadratic formula.

- Doctor Luis, The Math Forum
  http://mathforum.org/dr.math/