Reciprocals of Linear Functions MHF4U: Advanced Functions A linear - PDF document

r a t i o n a l f u n c t i o n s r a t i o n a l f u n c t i o n s Reciprocals of Linear Functions MHF4U: Advanced Functions A linear function has the form f ( x ) = kx + c for two real values k and c , such that k is the slope of the line and c is its y -intercept. The reciprocal of a linear function has the form 1 1 f ( x ) = kx + c , or f ( x ) = � . Reciprocals of Linear Functions x + c � k k Additional transformations may be applied to change the J. Garvin graph’s shape or position. J. Garvin — Reciprocals of Linear Functions Slide 1/19 Slide 2/19 r a t i o n a l f u n c t i o n s r a t i o n a l f u n c t i o n s Asymptotes Asymptotes The reciprocal of a linear function has two asymptotes: one The equation of a horizontal asymptote (HA) can be found vertical, and one horizontal. by dividing each term in a function by its highest power, then evaluating the function as x → ∞ . The vertical asymptote (VA) occurs for the value of x that 1 causes the denominator to equal zero. Consider the function f ( x ) = 5 x − 2. Since we are looking for x + c k = 0, the equation of the vertical asymptote is always x = − c 1 k . 1 x 5 x − 2 = 5 x x − 2 1 x For example, the function f ( x ) = 5 x − 2 will have a VA with 0 = equation x = 2 5 . 5 − 0 = 0 Thus, f ( x ) has a HA with equation f ( x ) = 0. This is true for all functions of this form. This technique will be handy later, so remember it. J. Garvin — Reciprocals of Linear Functions J. Garvin — Reciprocals of Linear Functions Slide 3/19 Slide 4/19 r a t i o n a l f u n c t i o n s r a t i o n a l f u n c t i o n s Asymptotes Asymptotes Example 1 Determine the equations of the asymptotes for f ( x ) = 2 x +7 , and state the domain and range. The VA has equation x = − 7 2 . The HA has equation f ( x ) = 0. −∞ , − 7 − 7 � � � � The domain is and the range is ∪ 2 , ∞ 2 ( −∞ , 0) ∪ (0 , ∞ ). J. Garvin — Reciprocals of Linear Functions J. Garvin — Reciprocals of Linear Functions Slide 5/19 Slide 6/19

r a t i o n a l f u n c t i o n s r a t i o n a l f u n c t i o n s Asymptotes Asymptotes Example 2 Determine the equations of the asymptotes for f ( x ) = 4 x − 9 , and state the domain and range. The VA has equation x = 9 4 . The HA has equation f ( x ) = 0. The 2 in the numerator stretches the graph, but does not change its location. � 9 � −∞ , 9 � � The domain is and the range is ∪ 4 , ∞ 4 ( −∞ , 0) ∪ (0 , ∞ ). J. Garvin — Reciprocals of Linear Functions J. Garvin — Reciprocals of Linear Functions Slide 7/19 Slide 8/19 r a t i o n a l f u n c t i o n s r a t i o n a l f u n c t i o n s Intercepts Intercepts As with any function, f ( x )-intercepts can be found by setting Example x = 0 in its equation. 1 Determine the intercepts for f ( x ) = x − 2. a Since a function of the form f ( x ) = kx + c has a horizontal asymptote at f ( x ) = 0, such functions have no x -intercepts. Evaluate f (0) to find the f ( x )-intercept. Vertical translations will shift a graph and its horizontal 1 asymptote, resulting in exactly one x -intercept. f (0) = 0 − 2 = − 1 2 Since there are no vertical translations, there are no x -intercepts. J. Garvin — Reciprocals of Linear Functions J. Garvin — Reciprocals of Linear Functions Slide 9/19 Slide 10/19 r a t i o n a l f u n c t i o n s r a t i o n a l f u n c t i o n s Intercepts Intercepts Example 1 Determine the intercepts for f ( x ) = x + 5 + 2. Evaluate f (0) to find the f ( x )-intercept. 1 f (0) = 0 + 5 + 2 = 11 5 To determine the x -intercept, set f ( x ) = 0 and solve for x . 1 0 = x + 5 + 2 − 2( x + 5) = 1 x + 5 = − 1 2 x = − 11 2 J. Garvin — Reciprocals of Linear Functions J. Garvin — Reciprocals of Linear Functions Slide 11/19 Slide 12/19

r a t i o n a l f u n c t i o n s r a t i o n a l f u n c t i o n s Intercepts Intervals of Increase and Decrease a Since there has been a vertical translation of 2 units up, the The graph of f ( x ) = kx + c is a hyperbola , with two horizontal asymptote has equation f ( x ) = 2. symmetric branches. There is a vertical asymptote at x = − 5. When a > 0, one branch is upper right and the other lower left. If a < 0, there has been a vertical reflection and the branches are in the lower right and upper left. Along with information about the asymptotes, this can be used to identify intervals of increase and decrease. J. Garvin — Reciprocals of Linear Functions J. Garvin — Reciprocals of Linear Functions Slide 13/19 Slide 14/19 r a t i o n a l f u n c t i o n s r a t i o n a l f u n c t i o n s Intervals of Increase and Decrease Intervals of Increase and Decrease � 1 � Example Since f ( x ) = − 4 , a < 0, so the function has x − 3 4 branches in the upper left and lower right. Given f ( x ) = − x − 3: As x → 3 from the left, f ( x ) → ∞ . Therefore, f ( x ) is • Determine the equations of the asymptotes increasing on ( −∞ , 3). • Determine the locations of the intercepts As x → ∞ , f ( x ) → 0 from below. Therefore, f ( x ) is • State the intervals on which the function is increasing or increasing on (3 , ∞ ). decreasing In general, the reciprocal of a linear function is always • State the intervals on which the slope is increasing or increasing when a < 0, and always decreasing when a > 0. decreasing There are asymptotes at x = 3 and f ( x ) = 0. The f ( x )-intercept is at 4 3 . There is no x -intercept. J. Garvin — Reciprocals of Linear Functions J. Garvin — Reciprocals of Linear Functions Slide 15/19 Slide 16/19 r a t i o n a l f u n c t i o n s r a t i o n a l f u n c t i o n s Intervals of Increase and Decrease Intervals of Increase and Decrease Since the left branch becomes steeper upwards as we move from left to right, the slope is increasing as x → 3 from the left. Since the right branch becomes less steep as we move from left to right, the slope is decreasing as x → ∞ . Therefore, the slope is increasing on ( −∞ , 3) and decreasing on (3 , ∞ ). It is important to remember the difference between the function increasing, and its slope increasing, as they are often different. J. Garvin — Reciprocals of Linear Functions J. Garvin — Reciprocals of Linear Functions Slide 17/19 Slide 18/19

r a t i o n a l f u n c t i o n s Questions? J. Garvin — Reciprocals of Linear Functions Slide 19/19