The General Polynomial FunctionDate: 09/17/98 at 22:55:29 From: prayoon Chaithaweep Subject: Pre-calculus Can you tell me how to solve these problems: 1) For the general polynomial function: n n-1 n-2 y = A x + A x + A x + ... + A 0 1 2 n what are the corresponding values of A , A , A , . . . A , and n in 0 1 2 n the linear function y = 6x - 2 ? 2) Write a linear function for A = 2/3, A = 7, and n = 1: 0 1 I know the answer of 2 is y = 2/3 + 7, but I don't know how to find it. Date: 09/18/98 at 12:25:10 From: Doctor Peterson Subject: Re: Pre-calculus Hello, Prayoon. I've had to reformat your question because the spacing got messed up, but I think I got it right. It's easier to write the subscripts as "A_0" for A-subscript-0, and "x^n" for x-to-the-nth power. They've made it a little tricky by using A_0 for the coefficient of the x^n term, but it's not too hard. First, n will be the highest exponent, so in problem 1, n = 1. Then you can just write the general form for n = 1, and write your given equation under it, remembering that x^0 = 1 and x^1 = x: y = A_0 x^1 + A_1 x^0 y = 6 x^1 - 2 x^0 So A_0 is 6 and A_1 is -2, since 6x - 2 can be written as: 6x^1 + (-2)x^0 The second problem just reverses the process. You are given n = 1, so you can write the generic form again, then fill in the values: y = A_0 x^1 + A_1 x^0 y = 2/3 x^1 + 7 x^0 = 2/3 x + 7 I suspect that your problem may be that you are not quite comfortable with what x^1 and x^0 are. I hope this helps you. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 09/18/98 at 17:20:51 From: Anonymous Subject: Re: Pre-calculus I still don't understand why n = 1 in problem 1. Also, how do I graph this function f(x) = square root 2x and how I find the inverse of it? Prayoon Date: 09/19/98 at 18:00:57 From: Doctor Peterson Subject: Re: Pre-calculus Hello, Prayoon. For further explanation of the first problem, in your formula for a polynomial, n is the highest exponent of x in the polynomial. In 6x - 2, which can be rewritten as 6x^1 - 2x^0, you have two powers of x, x = x^1 and 1 = x^0. So the highest exponent is 1. You want to graph y = sqrt(2x). You could just choose some values for x and plot them the way you make any graph, but let's approach this in a way that is connected with inverses. Your equation is the same as y^2 = 2x, or y^2/2 = x, with the restriction that y has to be positive, since the square root gives only the positive number whose square is 2x. So we can graph it by drawing the parabola x = y^2/2, looking at our paper sideways. Here's y = x^2 / 2, the same parabola going the other way: | * + * | + * | * ---+---+---*---+---+--- -2 -1 0 1 2 and here is x = y^2 / 2 (in which I have used only the positive values for y): | + * | + * | ---+---+---*---+---+--- -2 -1 0 1 2 To find the inverse, you have to solve for x in terms of y and state the restriction on the function's domain. That's really what we just did: x = y^2/2, with y positive. So the inverse of f is: f(y) = y^2/2 for y >= 0 (You could also write this as: f(x) = x^2/2 for x >= 0 since a function doesn't care what you call its variable.) I hope that gives you an idea of how inverses are related to graphs. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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