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### Algebra

```
Date: 5/10/96 at 10:26:55
From: Anonymous
Subject: Algebra

Find the domain, range, and zeros of

g(x) = x^2 - 6x + 4
```

```
Date: 10/17/96 at 12:51:17
From: Doctor Leigh
Subject: Re: Algebra

The domain of a function is all the values of x that can be put into
the function. In this case the domain is all real numbers.

The range of the function is all the values of Y that the function
covers. This function has a range of y >= -5.

The zeros of a function are found by setting the function equal to
zero and solving for x. The zeros of this function are

3 + sqrt(5)   and   3 - sqrt(5)

Use the quadratic equation.

-Doctor Leigh Ann,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```

```
Date: 09/01/2000 at 17:58:48
From: Tim Greene
Subject: Algebra

Hello Dr. Math:

I would like to expand upon the solution given for a problem in your
archives. In this problem, the writer states that the range of the
function x^2 - 6x + 4 is y >= -5. No method is shown for obtaining this
result. I assume the writer determined the function had a minimum value
when x = 3 and substituted that value of x in the function to find the
lower limit of the range for the function.

Of course calculus shows us quickly that the minimum value is when x = 3;
however, many high school students will not know calculus. The value of x
for which the function has a minimum, and that minimum value, can both be
obtained by completing the square in the given function; most second year
algebra students and many first year algebra students will be able to
understand this approach.

y = x^2 is a parabola opening upwards with vertex at (0,0);

y = x^2 + k is a parabola opening upwards with vertex at (0,k); and

y = (x-h)^2 + k is a parabola opening upwards with vertex at (h,k)

Therefore,

g(x) = x^2 - 6x + 4

= (x^2 - 6x + 9) - 5

= (x-3)^2 - 5

is a parabola with vertex (3,-5), so the range of the function is

[-5,infintiy]   or   y >= -5.

Tim Greene
```
Associated Topics:
High School Basic Algebra

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