Help!! Find the domain and range of y=(x^2) - 6x + 1?

Question

y=(x^2) - 6x + 1

In my summer review packet for calculus we have to find the domain and range of different functions by hand. I graphed this one and found the domain to be any real number, but i don't know how to find the range without using a calculator.

Answer 1

Complete the square to get it into vertex form.

(x^2-6x+9)-9+1

=(x-3)^2-8

R: {y=>-8, y ∈ R}

Answer 2

The domain is going to be the numbers that you can plug into x and get a meaningful value for y. Any real number will work for x.

The range is the numbers that you get for y when you put in any possible value for x. You can see that for large values of x you get large values of y. The values of y go up to infinity.

The question is, what is the minimum value for y? There are several ways of determining this. Graphing would help. Taking the derivative would help, but you would need to know some calculus to determine the derivative. Or a little trial and errror shows you that x=3 gives you the minimum value of y. At x=3 we have y = 3*3 - 6*3 + 1 = -8.

So the range is -8 to infinity.

Answer 3

Hi Jaymee,

This is taken from Dr. Math:

"The basic concept of domain and range is pretty simple. Here are the
facts:

- The domain of a function is the set of all the stuff you can
plug into the function.

- The range of a function is the set of all the stuff you can
get out of the function."


So, x is the domain, and y is the range. The value of y can approach positive infinity as the the values of x get arbitrarily large. But, does y have a minimum? Since you are taking calculus, we can take the derivative:

y' = 2·x - 6

setting y = 0, we get x = 3

Taking the derivative again, we get y'' = 2, which implies that that the critical point x = 3 is a minimum..

Now, we take the value x = 3, and plug it into the original equation to get 9 - 12 + 1 = -8.


So, the range is -8 <= y < infinity.

--------

I am grateful to Dr. Ivan for pointing out the error in my earlier answer, and his kind suggestions.

James :)
.

Answer 4

1. x -coordinate of vertex can be found using -b/(2a), so column 1 gives us -(-4)/(2*(-2)) = 4/(-4) = -1. Column 2 gives us -(-4)/(2*1) = 4/2 = 2. So column 2 is greater than column 1. 2. For any quadratic the domain is always all reals. 3. Find the vertex (see question 1). Substitute that in for x to determine y-coordinate. You get -6 for the x-coordinate and 39 for the y-coordinate. Since the lead term is negative, the parabola opens down and has a maximum at (-6,39) so the range is thus y ≤ 39.

Answer 5

You are correct in finding that the domain is the set of all reals.

For the range, notice that this is a parabola which opens up, so the range will be the set of all numbers >= the y-coordinate of the vertex. And you cab find the vertex by findind the minimum value of the function: y' = 2x - 6, and this vanishes when x = 3. At x = 3 we find y = -8, so the range is the set of reals >= -8.

Answer 6

The domain is the set of real numbers R.

To find the range, let us complete the square:
(x^2) - 6x + 1=(x^2) - 6x + 9 - 8=(x-3)^2 - 8
Since (x-3)^2 is always nonnegative, the range is
[-8,+infinity)

Answer 7

you should be able to have figured that out from the base of your parabola.
the range of a parabola where x^2 is a positive is any number greater than the vertex of the parabola.
you're going to need to do that sort of thing a LOT in calculus. (quartic functions are fun)