Explanation Given Information: The functions are ( i ) f ( x ) = 2 x + 3 at ( 1 , 5 ) . ( ii ) f ( x ) = x 2 + 4 at ( − 1 , 5 ) . ( iii ) f ( x ) = 5 x 2 at ( − 2 , 20 ) . ( iv ) f ( x ) = 2 x 2 − x at ( 2 , 6 ) . ( v ) f ( x ) = 2 x 2 − x + 5 at ( 0 , 5 ) . Graph: ( i ) Consider the function f ( x ) = 2 x + 3 and the point ( 1 , 5 ) Here, ( a , f ( a ) ) = ( 1 , 5 ) Now, compute the slope of the tangent line for the function f ( x ) = 2 x + 3 as follows: Put a = 1 , m tan = lim h → 0 f ( 1 + h ) − f ( 1 ) h To compute f ( 1 + h ) , substitute x = 1 + h in the function f ( x ) = 2 x + 3 . m tan = lim h → 0 [ 2 ( 1 + h ) + 3 ] − [ 2 ( 1 ) + 3 ] h Now, simplify 2 ( 1 + h ) by using the distributive property A ( B + C ) = A B + A C . m tan = lim h → 0 [ 2 + 2 h + 3 ] − 5 h Combine the like terms in the numerator. m tan = lim h → 0 2 h h Cancel out h and apply the limits. m tan = lim h → 0 2 = 2 Thus, the slope of the tangent to the graph of f ( x ) = 2 x + 3 at ( 1 , 5 ) is 2 . Now compute the slope intercept equation of the tangent line as follows: Here, the tangent line passes through the point ( 1 , 5 ) which implies x 1 = 1 and y 1 = 5 Substitute m = 2 , x 1 = 1 , y 1 = 5 in the equation y − y 1 = m ( x − x 1 ) . y − 5 = 2 ( x − 1 ) y − 5 = 2 x − 2 y = 2 x − 2 + 5 y = 2 x + 3 Now, graph the function and the tangent using the TI- 83 graphing calculator as follows: Step1. Press “ Y = ” key and insert the function and the tangent as, Y 1 = 2 X + 3 Y 2 = 2 X + 3 Step2. Press the “WINDOW” key and set the values as, X min = − 10 , Xmax = 10 and Xscl = 1 And Ymin = − 10 , Ymax = 10 and Yscl = 1 . Step3. Then press the “GRAPH” key. A window showing the graph of the function and the tangent will appear as, The graph of the function f ( x ) = 2 x + 3 is a line and there is no tangent drawn in the graph because a tangent line to a straight line is the line itself. ( ii ) Consider the function f ( x ) = x 2 + 4 and the point ( − 1 , 5 ) Here, ( a , f ( a ) ) = ( − 1 , 5 ) Now, compute the slope of the tangent line for the function f ( x ) = x 2 + 4 as follows: Put a = − 1 , m tan = lim h → 0 f ( − 1 + h ) − f ( − 1 ) h To compute f ( − 1 + h ) , substitute x = − 1 + h in the function f ( x ) = x 2 + 4 . m tan = lim h → 0 [ ( − 1 + h ) 2 + 4 ] − [ ( − 1 ) 2 + 4 ] h Now, simplify ( − 1 + h ) 2 by using the property ( A + B ) 2 = A 2 + 2 A B + B 2 . m tan = lim h → 0 [ 1 − 2 h + h 2 + 4 ] − 5 h Combine the like terms in the numerator. m tan = lim h → 0 h 2 − 2 h h Now, factor the numerator and divide the numerator as well as the denominator by h , m tan = lim h → 0 h ( h − 2 ) h = lim h → 0 ( h − 2 ) Apply the limits and simplify. m tan = 0 − 2 = − 2 Thus, the slope of the tangent to the graph of f ( x ) = x 2 + 4 at ( − 1 , 5 ) is − 2 . Now compute the slope intercept equation of the tangent line as follows: Here, the tangent line passes through the point ( − 1 , 5 ) which implies x 1 = − 1 and y 1 = 5 Substitute m = − 2 , x 1 = − 1 , y 1 = 5 in the equation y − y 1 = m ( x − x 1 ) . y − 5 = − 2 [ x − ( − 1 ) ] y − 5 = − 2 ( x + 1 ) y = − 2 x − 2 + 5 y = − 2 x + 3 Now, graph the function and the tangent using the TI- 83 graphing calculator as follows: Step1. Press “ Y = ” key and insert the function and the tangent as, Y 1 = X ^ 2 + 4 Y 2 = − 2X + 3 Step2. Press the “WINDOW” key and set the values as, X min = − 10 , Xmax = 10 and Xscl = 1 And Ymin = − 10 , Ymax = 10 and Yscl = 1 . Step3. Then press the “GRAPH” key. A window showing the graph of the function and tangent, will appear as, The line y = − 2 x + 3 is the tangent line to the graph of f ( x ) = x 2 + 4 at ( − 1 , 5 ) . ( iii ) Consider the function f ( x ) = 5 x 2 and the point ( − 2 , 20 ) Here, ( a , f ( a ) ) = ( − 2 , 20 ) Now, compute the slope of the tangent line for the function f ( x ) = 5 x 2 as follows: Put a = − 2 , m tan = lim h → 0 f ( − 2 + h ) − f ( − 2 ) h To compute f ( − 2 + h ) , substitute x = − 2 + h in the function f ( x ) = 5 x 2 . m tan = lim h → 0 [ 5 ( − 2 + h ) 2 ] − 5 ( − 2 ) 2 h Now, simplify ( − 2 + h ) 2 by using the property ( A − B ) 2 = A 2 − 2 A B + B 2 . m tan = lim h → 0 5 ( 4 − 4 h + h 2 ) − 20 h = lim h → 0 20 − 20 h + 5 h 2 − 20 h Combine the like terms in the numerator. m tan = lim h → 0 5 h 2 − 20 h h Now, factor the numerator and divide the numerator as well as the denominator by h , m tan = lim h → 0 5 h ( h − 4 ) h = lim h → 0 5 ( h − 4 ) Apply the limits and simplify

6th Edition

ISBN: 9780135963173

Author: Blitzer

Publisher: PEARSON CO

See similar textbooks

Concept explainers

Rate of Change

The relation between two quantities which displays how much greater one quantity is than another is called ratio.

Slope

The change in the vertical distances is known as the rise and the change in the horizontal distances is known as the run. So, the rise divided by run is nothing but a slope value. It is calculated with simple algebraic equations as:

Videos

Question

Chapter 11.4, Problem 60PE

To determine

To graph: Any five functions and their tangent lines without using the DRAW TANGENT feature of the graphing utility and check if the line appears to be tangent to the graph of f at the given point on f.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 11.4, Problem 59PE

Chapter 11.4, Problem 61PE

Students have asked these similar questions

16. Solve each of the following equations for x. (a) 42x+1 = 64 (b) 27-3815 (c) 92. 27² = 3-1 (d) log x + log(x - 21) = 2 (e) 3 = 14 (f) 2x+1 = 51-2x

11. Find the composition fog and gof for the following functions. 2 (a) f(x) = 2x+5, g(x) = x² 2 (b) f(x) = x²+x, g(x) = √√x 1 (c) f(x) = -1/2) 9 9(x) = х = - X

practice problem please help!

Chapter 11 Solutions

EP PRECALCULUS-MYLABMATH+ETEXT ACCESS

Ch. 11.1 - Check Point 1 Find: . Ch. 11.1 - Prob. 2CP Ch. 11.1 - Prob. 3CP Ch. 11.1 - Prob. 4CP Ch. 11.1 - Prob. 5CP Ch. 11.1 - Prob. 1CVC Ch. 11.1 - Prob. 2CVC Ch. 11.1 - Prob. 3CVC Ch. 11.1 - Fill in each blank so that the resulting statement...Ch. 11.1 - Fill in each blank so that the resulting statement...

Knowledge Booster

$Calculus$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.

To graph: Any five functions and their tangent lines without using the DRAW TANGENT feature of the graphing utility and check if the line appears to be tangent to the graph of f at the given point on f .

Solution Summary: The author explains how to graph any five functions and their tangent lines without using the DRAW TANGENT feature of the graphing utility.

Concept explainers

Videos

To graph: Any five functions and their tangent lines without using the DRAW TANGENT feature of the graphing utility and check if the line appears to be tangent to the graph of f at the given point on f.

Want to see the full answer?

Chapter 11 Solutions