Using x-Intercepts to Sketch Parabolas
Date: 08/04/2002 at 20:18:35 From: Beckie Subject: can't seem to get the right answers I've been working on a correspondence course over the summer and there are six questions where I cannot seem to get the answer that's in the back of the book. For each one, I'm supposed to sketch the graph of a parabola from an equation: a) y = 3x^2 -6x b) y = -2x^2 - 4x c) y = x^2 + 6x + 8 d) y = -x^2 + 4x + 5 e) y = 1/2x^2 + 2x -3 f) y = -x^2 -6x -8 The back of the book shows me the x-intercepts, but I'm not sure what that's supposed to tell me. Thanks, Beckie
Date: 08/04/2002 at 20:45:42 From: Doctor Ian Subject: Re: can't seem to get the right answers Hi Beckie, The x-intercepts are the points where the parabola intercepts the x-axis. At these points, the value of y is zero, so you can find them by setting y to zero and solving for x. For example, in y = 3x^2 - 6x we would set y = 0 to get 0 = 3x^2 - 6x = 3x(x - 2) Now, there are only two ways for this equation to be true, because the only way you can get a product of zero is for at least one of the factors to be zero. Does that make sense? In this case, if x is zero, the equation is true: 0 = 3(0)(0 - 2) It's also true if x is two: 0 = 3(2)(2 - 2) So the parabola intercepts the x-axis at the points (0,0) and (2,0). Because a parabola is symmetric, the vertex of the parabola has to be halfway between these points, right? That is, it will occur when x is 1. To find the value of y there, we substitute x = 1 into the original equation: y = 3x^2 - 6x = 3(1^2) - 6(1) = 3 - 6 = -3 So the vertex is at (1,-3). Now, this tells us something important. If the vertex is BELOW the x-axis, and if the parabola crosses the x-axis, then it has to open upward. Does that make sense? (If the vertex is ABOVE the x-axis, and if it crosses the x-axis, then it has to open downward. Note that a parabola doesn't have to cross the x-axis at all. But we won't worry about that for now.) So, from three points - (0,0), (1,-3), (2,0) - you can quickly sketch the graph of the equation. It may not be very accurate, except at those three points, but that's why it's called a 'sketch'. The first example was the easiest, because there was no constant term. An equation like y = x^2 + 6x + 8 is a little tougher, because you can't just pull an x out front. If you tried, you'd get y = x(x + 6 + 8/x) which wouldn't help you at all. So what can you do? Well, note that (x + 2)(x + 4) = x(x + 4) + 2(x + 4) = x^2 + 4x + 2x + 8 = x^2 + 6x + 8 Once you know this, you can go through the same drill. Set y to zero, 0 = (x + 2)(x + 4) This tells you that the x-intercepts must occur when x is equal to -2 and -4. So that tells you where the vertex is, and the rest is the same. Now, it's easy to go in this direction: (x + 2)(x + 4) --> x^2 + 6x + 8 You just apply the distributive property twice. But going in the other direction, x^2 + 6x + 8 --> (x + 2)(x + 4) is a little trickier. Essentially, you have to guess the answer, but there are some techniques that you can use to dramatically reduce the number of guesses that you have to try. You can read about how to attack these problems in the following items from our archive: Factoring Polynomials http://mathforum.org/library/drmath/view/60580.html Factoring Quadratics Without Guessing http://mathforum.org/library/drmath/view/60700.html I think they'll give you a little more insight than your textbook. Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/
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