Intersection of a Parabola and a LineDate: 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/ |
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