Consider the function f ( x ) = 1 x - 1 + 2 . Sketch the graph of f ( x ) . Show the asymptotes as dotted lines.

Given function is f ( x ) = 1 x - 1 + 2 Comparing above equation with f ( x ) = p c x + d + q , we have c = 1 , d = - 1 , q = 2 The horizontal asymptote is y = q = 2 The vertical asymptote is x = - d c = 1 To sketch the graph of y = f x = 1 x - 1 + 2 : x 0 1 2 3 4 5 y 1 undefined 3 2 . 5 2 . 33 2 . 25 Similarly consider negative values of x and find y and use those points to sketch the graph

Consider the function f ( x ) = 1 x - 1 + 2 . Sketch the graph of f ( x ) . Show the asymptotes as dotted lines.

Given function is f ( x ) = 1 x - 1 + 2 Comparing above equation with f ( x ) = p c x + d + q , we have c = 1 , d = - 1 , q = 2 The horizontal asymptote is y = q = 2 The vertical asymptote is x = - d c = 1 To sketch the graph of y = f x = 1 x - 1 + 2 : x 0 1 2 3 4 5 y 1 undefined 3 2 . 5 2 . 33 2 . 25 Similarly consider negative values of x and find y and use those points to sketch the graph

Use your GDC to solve given equation graphically. x + 3 2 x - 1 = 4 - x

Let us consider f x = x + 3 2 x - 1 and g x = 4 - x To sketch the graph of f x = x + 3 2 x - 1 : x - 4 - 3 - 2 - 1 0 1 2 3 4 f x 0 . 11 0 - 0 . 2 - 0 . 66 - 3 4 1 . 667 1 . 2 1 To sketch the graph of g x = 4 - x : x - 4 - 3 - 2 - 1 0 1 2 3 4 g x 8 7 6 5 4 3 2 1 0 Using the above points plot the graph of f x and g x From the graph it is clear that f x and g x intersect at x = 1 . 293 and x = 2 . 707 .

Use your GDC to solve given equation graphically. 10 x + 3 2 x + 4 = 4 - x 2

Let us consider f x = 10 x + 3 2 x + 4 and g x = 4 - x 2 To sketch the graph of f x = 10 x + 3 2 x + 4 : x - 4 - 3 - 2 - 1 0 1 2 3 4 f x 9 . 25 13 . 5 undefined - 3 . 5 0 . 75 2 . 166 2 . 875 3 . 3 3 . 58 To sketch the graph of g x = 4 - x 2 : x - 4 - 3 - 2 - 1 0 1 2 3 4 g x - 12 - 5 0 3 4 3 0 - 5 - 12 Using the above points plot the graph of f x and g x From the graph it is clear that f x and g x intersect at x = 1 . 266 .

Let f x = 2 x - 8 1 - x . Find the x - intercept.

Let the function be y = f x = 2 x - 8 1 - x The axial intercepts are the points where the graph crosses the axes (both the x - axis and y - axis). They are called the x - intercept and the y - intercept respectively. To find the x - intercept substitute y = 0 : 2 x - 8 1 - x = 0 ⇒ 2 x - 8 = 0 ⇒ x = 4 Therefore, x - intercept is 4 , 0 .

The function f x is defined as f x = 4 + 1 2 x - 5 . Sketch the curve of f for - 2 ≤ x ≤ 5 .

Given function is f x = 4 + 1 2 x - 5 ⇒ y = 4 + 1 2 x - 5 Comparing above equation with f ( x ) = p c x + d + q , we have c = 2 , d = - 5 , q = 4 The horizontal asymptote is y = q = 4 The vertical asymptote is x = - d c = 5 2 = 2 . 5 Now, we can draw the curve of f using the asymptotes:

E M B I B E

Mathematics : Analysis and Approaches Standard Level Course Companion>Chapter 4 - Equivalent Representations: Rational Functions>Chapter review>Q 3

MEDIUM

Diploma

IMPORTANT

Earn 100

Consider the function $f (x) = \frac{1}{x - 1} + 2$ . Sketch the graph of $f (x)$ . Show the asymptotes as dotted lines.

Important Questions on Equivalent Representations: Rational Functions

HARD

Diploma

IMPORTANT

Mathematics : Analysis and Approaches Standard Level Course Companion>Chapter 4 - Equivalent Representations: Rational Functions>Chapter review>Q 4

Use your $GDC$ to plot the graphs of $f (x) = \frac{2 x + 1}{2 x - 1}$ and $g (x) = x + 2$ on the same set of axes. By considering the point of intersection, solve the equation $\frac{2 x + 1}{2 x - 1} = x + 2$ .