NONLINEAR FUNCTIONS A. Absolute Value Exercises: 2. We need to scale the graph of Qx ( )

Transcription

1 NONLINEAR FUNCTIONS A. Absolute Value Eercises:. We need to scale the graph of Q ( ) f ( ) =. The graph is given below. = by the factor of to get the graph of We need to scale the graph of Q ( ) f ( ) =. The graph is given below. = by the factor of to get the graph of

2 . To get the graph of f( ) = by using Q ( ) =, firstly scale Q ( ) = by a factor of and then shift the graph unit in the direction To draw the graph of G ( ) = + by using Q ( ) =, firstly reflect the graph of Q ( ) = across the ais and then shift the new graph unit in the direction The graph of R ( ) = + can be obtained from Q ( ) = by Scaling the graph of Q ( ) = by a factor of / Shifting the new graph unit in direction and units in y direction.

3 The graph of T( ) = + can be obtained from Q ( ) = by shifting the graph of Q ( ) = first unit in direction and then units in y direction The graph of H( ) = = = can be obtained from Q ( ) = by Scaling the graph of Q ( ) = by a factor of Shifting the new graph / units in direction.

4 The graph of ( ) H( ) = + = + = + can be obtained from Q ( ) = by Scaling the graph of Q ( ) = by a factor of Shifting the new graph units in direction and unit in y direction ( + ), if +, if + 9. G ( ) = + = = ( ( + )), if + < +, if + < +. Hence, G ( ) can be written as:, if G ( ) =. +, if < and

5 ( + ), if +, if +. T( ) = + = = ( + ), if + <, if + < +. Then,, if T( ) =., if < and ( ), if, if + +. R ( ) = + = = 5 ( ) +, if < +, if <. So, +, if R ( ) =. 5 +, if < and, if, if. H( ) = = = ( ), if < +, if <. Thus,, if / H( ) =. +, if < / and (+ ), if + +, if +. M( ) = + = = (+ ), if + < 5, if + < +. Hence, +, if M( ) =. 5, if < and B. Polynomial Functions Eercises:. P ( ) = + 8= is a polynomial of degree.... = + + = is a polynomial of degree. F( ) = + = + + is a second degree polynomial f ( ) a b c 5 5 = + + = is a polynomial of degree 5. h ( )

6 5. To graph the polynomial function f ( ) = +, we need to reflect the graph of f( ) = across the ais and then shift the new graph unit in y direction To graph g ( ) = +, we need to Reflect the graph of f ( ) = across the ais Scale the new graph by a factor of Shift the graph units in y direction

7 . To draw the graph of h ( ) = ( ) +, we need to: Reflect the graph of f ( ) = across the ais Scale the new graph by a factor of Shift the graph unit in direction and unit in y direction To draw the graph of F( ) = ( ), we need to: Scale the graph of f ( ) = by a factor of Shift the graph units in direction and units in y direction.

8 To obtain the graph of F( ) = ( + ) from the graph of Reflect the graph of g( ) = across the ais g( ) = ; Scale the graph by a factor of Shift the new graph units in direction and units in y direction.. To obtain the graph of Scale the graph of = + from the graph of F( ) ( ) 5 g( ) = by a factor of g( ) = ; Shift the new graph units in direction and 5 units in y direction.. To obtain the graph of F( ) = (+ ) + = + + = + + from the graph of g( ) = ; Scale the graph of g( ) = by a factor of Shift the new graph / units in direction and units in y direction. C. Rational Functions Eercises:

9 . f( ) = has vertical asymptotes when = = =± (Notice that the numerator is not when =± ). Since the degree of the numerator is smaller than the degree of the denominator, the horizontal asymptote is y =. +. The vertical asymptote of g ( ) = is = since = = (the numerator is not zero at this value). The degrees of the numerator and the denominator are equal, so y = = is the + horizontal asymptote of g ( ) =. +. To find the vertical asymptote(s) of h ( ) = we need to solve the equation. + + =. We have: + = ( )( ) = = or =. Since the numerator is not at these values of, the vertical asymptotes are = and =. The numerator and the denominator are same degree polynomials; horizontal asymptote. y = = is the +. R ( ) = has vertical asymptotes if + = has solutions. Since + + = ( + )( ) = = or =, the vertical asymptotes are = and = (the numerator is not for these values). The horizontal asymptote is y = since the degree of the numerator is smaller than the degree of the denominator. 5. i) f( ) = ; Domain: all values of ecept = ±. The intercept is = since f () =. The y intercept is y =. + ii) g ( ) = ; Domain: all values of ecept =. The intercept is = / since g( /) =. The y intercept is y = / since g () = /.

10 + iii) h ( ) = + ; Domain: all values of ecept = and =. There is no intercepts since + = has no solutions. The y intercept is y = / since f () = /. + iv) R ( ) = ; + Domain: all values of ecept = and =. The intercept is = / since R( /) =. The y intercept is y = / since f () = /.. The graph of h ( ) = is given below The vertical asymptote of h ( ) = is = and the horizontal asymptote is y =. To get the graph of h ( ) = from g ( ) =, we need to: Reflect the graph of g ( ) = across the ais Scale the new graph by a factor of Shift the graph units in direction.

11 . The graph of + g ( ) = is given below The vertical asymptote of g ( ) = is = and the horizontal asymptote is y =. + To get the graph of g ( ) = = + from R ( ) =, we need to: Scale the graph of R ( ) = by a factor of Shift the graph units in direction and units in y direction. 8. The graph of + f( ) = is given below.

12 The vertical asymptote of f( ) = is = and the horizontal asymptote is y =. + To get the graph of f( ) = = + from g ( ) =, we need to: Scale the graph of g ( ) = by a factor of Shift the graph unit in direction and units in y direction. D. Eponential Functions Eercises:. The horizontal asymptote for the graph of g= ( ) is y =. The y intercept is y = since g () = = =.. We have ( ) ( ) ( ) ( )( ) f = = = =, which means that f( ) = have the same graph. f( ) = and

13 . The horizontal asymptote for the graph of f ( ) y = 5. The y intercept is y = since g () = + 5 =. ( ) = + 5= + 5= + 5 is. To get the graph of g= ( ) from f( ) =, we need to shift the graph of f( ) = units in y direction. 5. To get the graph of + g ( ) = f( ) = units in direction units in y direction. from f( ) =, we need to shift the graph of. To get the graph of ( ) g= ( ) 5 = 5 = 5 from f( ) = 5, we need to shift the graph of f( ) = 5 unit in y direction.. To graph g= ( ), shift the graph of f( ) = units in y direction To get the graph of g ( ) = we need to shift the graph of units in direction and units in y direction. f( ) =

14 The graph of g ( ) = + is obtained by shifting f( ) = units in y direction

15 E. The number e, Radioactive Decay, and savings accounts Eercises: t t. To draw the graph of pt () = e + we need to shift the graph of f () t = e unit in y direction To draw the graph of gt () = e t + we need to reflect the graph of f () t the y ais and then shift it units in y direction. t = e across The graph of.t = e is given below. f() t

16 t (measured in years) Percentage remaining.5t e.9859 = 98.5%.89 = 8.% 5.55 =.%. =.% =.5% 5. t (measured in years) Percentage remaining.5t e.9955 = 99.55% = 95.59% = 9.85%.85 =.% =.5%. We want to estimate the time when the percentage remaining is 5%. If you look at the table in Eercise-, you ll see that after 5 years.% of the substance is left. So, ½ life of this substance must be smaller than 5.

17 .5t Percentage remaining is given by e.5t. Thus, we need to find t such that e =.5. If we take the natural logarithms of both sides, we get.5t = ln(.5). That is, ln(.5) t = = Thus, the / life of this substance is approimately. years. rt. We know that A dollars will be worth At () = Ae dollars after t years. Here, A =, r =.8 and t =. Since (.8)().8 A() = e = e = 5.5, there will be 5.5 dollars in the account after years. rt 8. In the formula At () = Ae, A =, r =. and t =. We have (.)() A() = e = e = 9.. Thus, there is approimately 9,. dollars in the account.