What Are Quadratic Functions?Date: 02/27/2003 at 23:14:16 From: Fly Subject: Quadratic Functions What are quadratic Functions? I do not know the difference between a quadratic function and a quadratic formula. Date: 03/03/2003 at 09:47:59 From: Doctor Ian Subject: Re: Quadratic Functions Hi Fly, A quadratic function looks like this: f(x) = ax^2 + bx + c where a, b, and c are constants. When the value of the function is zero (i.e., where the graph of the function crosses the x-axis), you have a quadratic equation in standard form: 0 = ax^2 + bx + c When you have a quadratic equation in standard form, you can use the quadratic formula to find the values of x that satisfy it. These values are called the 'roots' of the function. An example might help. The quadratic function f(x) = x^2 + 3x - 4 describes a parabola. This parabola crosses the x-axis where 0 = x^2 + 3x - 4 The values of x at those points are given by the quadratic formula, -b +/ sqrt(b^2 - 4ac) x = --------------------- 2a where a, b, and c are the coefficients of the quadratic, linear, and constant terms. In this case, a=1, b=3, and c=-4. So -(3) +/ sqrt((3)^2 - 4(1)(-4)) x = ------------------------------ 2(1) Note that I used parentheses for the substitutions, because it helps me keep track of the signs. Getting the signs mixed up is very easy to do with this formula, and can give you very strange results. Anyway, I can simplify this: -(3) +/ sqrt((3)^2 - 4(1)(-4)) x = ------------------------------ 2(1) -3 +/ sqrt(9 - -16) = ------------------ 2 -3 +/ sqrt(25) = -------------- 2 -3 +/ 5 = ------- 2 = (-3 + 5)/2 or (-3 - 5)/2 = 1 or -4 So x = 1 and x = -4 are the roots of the function. We can check this by evaluating the function at these values: f(1) = (1)^2 + 3(1) - 4 = 1 + 3 - 4 = 0 and f(-4) = (-4)^2 + 3(-4) - 4 = 16 - 12 - 4 = 0 To sum up, when we have a quadratic formula, it will give us a value corresponding to any input value that we choose for x. If we'd like to graph that function, the easiest way to do that is to find where it crosses the x-axis. That gives us two points, and from symmetry, we know that the vertex of the parabola has to be halfway between those points. So that gives us three points, which is good enough for a quick sketch. To find the x-intercepts, we set the value of the quadratic _function_ to zero, and use the quadratic _formula_ to find the values of x that make the resulting quadratic _equation_ true. Does that make sense? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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