Given the circle (x+2)^2 + (y-1)^2 = 25, write an equation for the function f whose graph is in the upper half of this circle. give the domain and range of the function f expressed in interval notation. * show work *

 f(x) = 1 + √(- x² - 4x + 21), D: [-7, 3], R: [1, 6]

Explanation:

(x + 2)² + (y - 1)² = 25   ⇒ center (h, k) = (-2, 1) and radius (r) = 5

Equation of a semicircle is to solve the equation for y.  

(x + 2)² + (y - 1)² = 25

               (y - 1)² = 25 - (x + 2)²

               (y - 1)² = 25 - (x² + 4x + 4)

               (y - 1)² = 25 - x² - 4x - 4

               (y - 1)² = - x² - 4x + 21

              √(y - 1)² = √(- x² - 4x + 21)

                  y - 1   = +/- √(- x² - 4x + 21)

                     y     = 1 +/- √(- x² - 4x + 21)

Since we are looking for the top half, the equation is:

                       y = 1 + √(- x² - 4x + 21)

in function format:

                     f(x) = 1 + √(- x² - 4x + 21)

NOTE: The easiest way to find the domain and range is to graph the circle.  Below is how to find the domain and range algebraically.

Domain: square root must be greater than or equal to zero

- x² - 4x + 21 ≥ 0

 x² + 4x - 21 ≤ 0  divided both sides by -1

(x + 7)(x - 3) = 0   factored to find the zeros

 x + 7 = 0      x - 3 = 0    

     x = -7         x = 3

The domain must be between -7 and 3 inclusively ; -7 ≤ x ≤ 3.

Interval Notation: D: [-7, 3]

Range: find the y-value at one of the zeros (x = -7 or x = 3) and the y-value of the vertex (the axis of symmetry goes through the center, x = -2)

f(x) = 1 + √(- x² - 4x + 21)

f(-7) = 1 + √(- (-7)² - 4(-7) + 21)

       = 1 + √(-49 + 28 + 21)

       = 1 + √(0)

        = 1

f(-2) = 1 + √(- (-2)² - 4(-2) + 21)

       = 1 + √(-4 +8 + 21)

       = 1 + √(25)

       = 1 + 5

       = 6

the range must be between 1 and 6 inclusively ; 1 ≤ y ≤ 6

Interval Notation: R: [1, 6]

okay so 15.15 *()5: +5/6

step-by-step explanation:

your answer is the frist one

hop i