For the function below: State the domain and range. y = 1 x + 2 + 5 .

The given function is y = 1 x + 2 + 5 . The equation of the horizontal asymptote is y = k and the equation of the vertical asymptote is x = h . Now on comparing f x = 1 x + 2 + 5 with f x = a x - h + k . We get, h = - 2 and k = 5 . So, x = h = - 2 and y = k = 5 . The domain of the rational function f x = a x - h + k is x ∈ R , x ≠ h . Here, x ≠ - 2 . The range of the rational function f x = a x - h + k is y ∈ R , y ≠ k . Here, y ≠ 5 . Hence, the domain is x ≠ - 2 and the range is y ≠ 5 .

For the function below: State the domain and range. y = 1 x - 3 + 1 .

The given function is y = 1 x - 3 + 1 . The equation of the horizontal asymptote is y = k and the equation of the vertical asymptote is x = h . Now on comparing y = 1 x - 3 + 1 with f x = a x - h + k . We get, h = 3 and k = 1 . So, x = h = 3 and y = k = 1 . The domain of the rational function f x = a x - h + k is x ∈ R , x ≠ h . Here, x ≠ 3 . The range of the rational function f x = a x - h + k is y ∈ R , y ≠ k . Here, y ≠ 1 . Hence, the domain is x ≠ 3 and the range is y ≠ 1 .

Describe the individual transformations on f x = 1 x to obtain f x = 1 x - h + k for set of values for h and k . h > 0 , k = 0 .

We need describe the individual transformations on f x = 1 x to obtain f x = 1 x - h + k for set of values for h and k . Here, h > 0 , k = 0 . We know that x = h and y = k Since, h > 0 , f x = 1 x - h + k is always in the first or third quadrant reflects on x - axis. Hence, the transformation in the first or third quadrant reflects on x - axis.

Describe the individual transformations on f x = 1 x to obtain f x = 1 x - h + k for set of values for h and k . k > 0 , h = 0 .

We need to describe the individual transformations on f x = 1 x to obtain f x = 1 x - h + k for set of values for h and k . Here, k > 0 , h = 0 . We know that x = h and y = k Since, k > 0 , f x = 1 x - h + k is always in the first or second quadrant reflects on y - axis. Hence, the transformation in the first or second quadrant reflects on y - axis.

Describe the individual transformations on f x = 1 x to obtain f x = 1 x - h + k for set of values for h and k . h < 0 , k = 0 .

We need to describe the individual transformations on f x = 1 x to obtain f x = 1 x - h + k for set of values for h and k . Here, h < 0 , k = 0 . We know that x = h and y = k . Since, h < 0 , f x = 1 x - h + k is always in the second or third quadrant reflects on x - axis. Hence, the transformation in the second or third quadrant reflects on x - axis.

Describe the individual transformations on f x = 1 x to obtain f x = 1 x - h + k for set of values for h and k . k < 0 , h = 0 .

We need to describe the individual transformations on f x = 1 x to obtain f x = 1 x - h + k for set of values for h and k . Here, k < 0 , h = 0 . We know that x = h and y = k . Since, k < 0 , f x = 1 x - h + k is always in the third or fourth quadrant reflects on y - axis. Hence, the transformation in the third or fourth quadrant reflects on y - axis.

E M B I B E

Extended MYP Mathematics A Concept based approach Years 4 & 5>What Comes Next?>Chapter 12 - Graphs of Rational Functions>Practice 2

Rose Harrison, Clara Huizink, Aidan Sproat Clements and, Marlene Torres Skoumal Solutions for Exercise 16: Practice 2

Author:Rose Harrison, Clara Huizink, Aidan Sproat Clements & Marlene Torres Skoumal

Rose Harrison Mathematics Solutions for Exercise - Rose Harrison, Clara Huizink, Aidan Sproat Clements and, Marlene Torres Skoumal Solutions for Exercise 16: Practice 2

Attempt the free practice questions from Exercise 16: Practice 2 with hints and solutions to strengthen your understanding. Extended MYP Mathematics A Concept based approach Years 4 & 5 solutions are prepared by Experienced Embibe Experts.