Find the domain and range of the function

f(x)=x^2-12x+2

y = f(x) = x^2-12x +2

Let's complete the square to see what this parabola looks like
x^2-12 x = y-2
x^2-12 x +36 = y+34
(x-6)^2 = y+34
Well, this parabola is symmetric about the vertical line where x = 6
As x gets big positive or negative, y gets big so this parabola is upright, holds water.
So the domain of y is all real x.
The left side is always positive, so y+34 never gets less than zero, so y never gets less than -34
so the range of y is all real y greater than or equal to -34

Typo - I mean the domain of x is all real x

thanks

To find the domain and range of a function, we need to understand the concept of domain and range.

Domain refers to the set of all possible input values (or x-values) for a function. In other words, it represents the values of x for which the function is defined.

Range, on the other hand, refers to the set of all possible output values (or y-values) for a function. In other words, it represents the values of y that the function can take.

For the function f(x) = x^2 - 12x + 2, there are no specific restrictions on the input values because this is a quadratic function.

To find the domain, we can assume that any real number can be input into the function. So, the domain is (-∞, ∞), which means all real numbers.

To find the range, we can use a few different strategies. One approach is to analyze the graph of the function. However, it's important to note that this function is a quadratic function, and its graph forms a parabola. A parabola either opens upward or downward depending on the coefficient of the squared term, which in this case is positive. This means that the parabola opens upward, and therefore there is no lower bound to the range.

To find the upper bound of the range, we can consider the vertex of the parabola. The x-coordinate of the vertex can be found using the formula x = -b/2a, where a is the coefficient of the squared term and b is the coefficient of the linear term. In this case, a = 1 and b = -12.

x = -(-12)/(2*1) = 12/2 = 6

Now, we can substitute the x-coordinate of the vertex into the function to find the corresponding y-coordinate.

f(6) = 6^2 - 12(6) + 2 = 36 - 72 + 2 = -34

Therefore, the vertex of the parabola is (6, -34). Since the parabola opens upward and the vertex is the lowest point on the graph, the range of the function is (-34, ∞).

In summary, the domain of the function f(x) = x^2 - 12x + 2 is (-∞, ∞) and the range is (-34, ∞).