How do you find the domain and range of #f(x)=x^2-12x+36#?

Question

Ethan Adkins · Accepted Answer

Find domain and range of #f(x) = x^2 - 12x + 36#

#f(x) = ( x - 6)^2#

The parabola opens upward. There is a Min at vertex #(6, 0)#.

There is also double root at #x = 6#

Domain of #x#: #(-oo, + oo)#

Range of #y#: #(-oo, + oo)#

plot{(x - 6)^2 [-10, 10, -5, 5]}

Nathan Axel · Answer

undefined To find the domain and range of the function $ f(x) = x^2 - 12x + 36 $:

1. **Domain**: Since the function is a polynomial, the domain is all real numbers.

2. **Range**: To find the range, we need to analyze the graph of the function. The function is a quadratic function, and its graph is a parabola opening upwards. Since the coefficient of $ x^2 $ is positive, the parabola opens upwards and has a minimum value. The minimum value of the function occurs at the vertex of the parabola.

To find the vertex, we use the formula $ x = -\frac{b}{2a} $, where $ a = 1 $ (coefficient of $ x^2 $) and $ b = -12 $ (coefficient of $ x $). Plugging in the values, we get $ x = \frac{12}{2} = 6 $.

Substituting $ x = 6 $ into the function, we get $ f(6) = 6^2 - 12(6) + 36 = 36 - 72 + 36 = 0 $.

So, the minimum value of the function is $ f(6) = 0 $.

Since the parabola opens upwards and has a minimum value of 0, the range of the function is all real numbers greater than or equal to 0, i.e., $ \text{Range} = \{ y \in \mathbb{R} \mid y \geq 0 \} $.