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```Date: 02/27/2003 at 23:14:16
From: Fly

I do not know the difference between a quadratic function and a
```

```
Date: 03/03/2003 at 09:47:59
From: Doctor Ian

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/
```
Associated Topics:
High School Basic Algebra
High School Definitions
High School Functions

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