Question

y=x^2-4x+4
A. miniumum or maximum
B. State the maximum or minimum values.
C. What are the domain and range

Answer 1

Is this calculus or algebra? I know of two ways to work with this kind of problem.

Answer 2

Intermediate Algebra

Answer 3

Okay. For a parabola of the form
\[f(x)=ax^2+bx+c, \;(a\not=0),\]
the graph of f(x) looks like one of the two following parabolas:

So, when a > 0, the parabola will have a minimum at its vertex; if a < 0, the parabola will have a maximum at its vertex.

Answer 4

To find the actual min/max value, I believe the method is to complete the square and write the parabola in vertex form (unless you learned some other method).
\[x^2-4x+4\\
(x-2)^2\]
This tells you that the vertex is (2, 0).

(Recall that vertex form is the following:
\[f(x)=a(x-h)^2+k, \text{ with }(h,k) \text{ being the vertex.)}\]

Answer 5

The domain is the set of all numbers for which f(x) is defined. In this case, you can put any number into f(x) and you'll get another number, so the domain is all real numbers. This is true for all polynomials.

As for range, that depends on the max/min value of the parabola. For example, consider the parabola
\[y=x^2\]
It has a vertex at (0,0).
So, the range of y = x^2 is the set of all non-negative integers.