Calculus: Graphical, Numerical, Algebraic

3rd Edition

ISBN: 9780133688399

Author: Finney, Ross L.

Publisher: Pearson/Prentice Hall

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Question

Chapter 2.4, Problem 12E

(a)

To determine

To find: The slope of the curve y=x2−3x−1 at x=0.

(a)

Expert Solution

Answer to Problem 12E

The slope of the curve y=x2−3x−1 at x=0 is −3.

Explanation of Solution

Given information:

The curve is y=x2−3x−1 and the point is x=0.

Calculation:

The formula for the slope of the curve f(x) at (a,f(a)) is equal to limh→0f(a+h)−f(a)h.

The slope of the curve y=x2−3x−1 at x=0 is:

Slope=limh→0f(0+h)−f(0)h=limh→0f(h)−f(0)h

Substitute h for x in the function.

f(x)=x2−3x−1f(h)=h2−3h−1

Substitute 0 for x in the function.

f(x)=x2−3x−1f(0)=02−3×0−1f(0)=0−0−1f(0)=−1

Substitute h2−3h−1 for f(h) and −1 for f(0) in the formula for slope.

Slope=limh→0h2−3h−1−(−1)h=limh→0h(h−3)h=limh→0(h−3)=−3

Therefore, the slope of the curve y=x2−3x−1 at x=0 is −3.

(b)

To determine

To find: The equation of the tangent line to the curve y=x2−3x−1 at x=0.

(b)

Expert Solution

Answer to Problem 12E

The equation of the tangent line to the curve y=x2−3x−1 at x=0 is y=−(3x+1).

Explanation of Solution

Given information:

The curve is y=x2−3x−1 and the point is x=0.

Calculation:

The equation of the tangent line to

y=f(x) at (a,f(a)) with slope at x=a is:

y−f(a)=m(x−a)

The function is y=x2−3x−1 and as calculated in part (a), the value of f(0) is equal to −1 . The value of slope of tangent to curve y=x2−3x−1 at x=0 is −3.

Substitute −3 for m , 0 for a and −1 for f(0) in the equation of tangent.

y−f(0)=−3(x−0)y−(−1)=−3xy+1=−3xy=−(3x+1)

Therefore, the equation of the tangent line to the curve y=x2−3x−1 at x=0 is y=−(3x+1).

(c)

To determine

To find: The equation to the normal line of the curve y=x2−3x−1 at x=0.

(c)

Expert Solution

Answer to Problem 12E

The equation of the normal line to the curve y=x2−3x−1 at x=0 is y=13x−1.

Explanation of Solution

Given information:

The curve is y=x2−3x−1 and the point is x=0.

Calculation:

The equation of the normal line of y=f(x) at (a,f(a)) with slope at x=a is:

y−f(a)=m(x−a)

The function is y=x2−3x−1 and as calculated in part (a), the value of f(0) is −1 . The value of slope of tangent to curve y=x2−3x−1 at x=0 is −3.

So, the slope of the normal to the curve y=x2−3x−1 at x=0 is 13.

Substitute 13 for m , 0 for a and −1 for f(0) in the equation of normal.

y−f(0)=13(x−0)y−(−1)=13xy+1=13xy=13x−1

Therefore, the equation of the normal line to the curve y=x2−3x−1 at x=0 is y=13x−1.

(d)

To determine

To graph: The curve y=x2−3x−1 , the tangent and normal line to the curve at x=0 in the same square viewing window.

(d)

Expert Solution

Explanation of Solution

Given information:

The curve is y=x2−3x−1 and the point is x=0.

Graph:

As calculated in part (b), the equation of the tangent line to the curve y=x2−3x−1 at x=0 is y=−(3x+1) . As calculated in part (c), the equation of normal line to the curve y=x2−3x−1 at x=0 is y=13x−1.

To graph the curve, tangent and normal to the curve, follow the steps using graphing calculator.

First press “ON” button on graphical calculator, press Y= key and enter right hand side of the equation y=x2−3x−1 after the symbol Y1 . Enter the keystrokes X^2−3X−1 . Now press the ENTER button and enter right hand side of the equation y=−(3x+1) after the symbol Y2 . Enter the keystrokes (−)(3X+1) . Now press the ENTER button and enter right hand side of the equation y=13x−1 after the symbol Y3 . Enter the keystrokes ((1÷3)X)−1.

The display will show the equations,

Y1=X^2−3X−1Y2=−(3X+1)Y3=((1/3)X)−1

Now, press the GRAPH key and TRACE key to produce the graph of given function in standard window as shown below.

Calculus: Graphical, Numerical, Algebraic, Chapter 2.4, Problem 12E

Figure (1)

Interpretation: From the graph it can be seen that the curve, the tangent and the normal line intersect at the point (0, −1).

Chapter 2.4, Problem 11E

Chapter 2.4, Problem 13E

Chapter 2 Solutions

Calculus: Graphical, Numerical, Algebraic

Ch. 2.1 - Prob. 1QR Ch. 2.1 - Prob. 2QR Ch. 2.1 - Prob. 3QR Ch. 2.1 - Prob. 4QR Ch. 2.1 - Prob. 5QR Ch. 2.1 - Prob. 6QR Ch. 2.1 - Prob. 7QR Ch. 2.1 - Prob. 8QR Ch. 2.1 - Prob. 9QR Ch. 2.1 - Prob. 10QR

Ch. 2.1 - Prob. 1E Ch. 2.1 - Prob. 2E Ch. 2.1 - Prob. 3E Ch. 2.1 - Prob. 4E Ch. 2.1 - Prob. 5E Ch. 2.1 - Prob. 6E Ch. 2.1 - Prob. 7E Ch. 2.1 - Prob. 8E Ch. 2.1 - Prob. 9E Ch. 2.1 - Prob. 10E Ch. 2.1 - Prob. 11E Ch. 2.1 - Prob. 12E Ch. 2.1 - Prob. 13E Ch. 2.1 - Prob. 14E Ch. 2.1 - Prob. 15E Ch. 2.1 - Prob. 16E Ch. 2.1 - Prob. 17E Ch. 2.1 - Prob. 18E Ch. 2.1 - Prob. 19E Ch. 2.1 - Prob. 20E Ch. 2.1 - Prob. 21E Ch. 2.1 - Prob. 22E Ch. 2.1 - Prob. 23E Ch. 2.1 - Prob. 24E Ch. 2.1 - Prob. 25E Ch. 2.1 - Prob. 26E Ch. 2.1 - Prob. 27E Ch. 2.1 - Prob. 28E Ch. 2.1 - Prob. 29E Ch. 2.1 - Prob. 30E Ch. 2.1 - Prob. 31E Ch. 2.1 - Prob. 32E Ch. 2.1 - Prob. 33E Ch. 2.1 - Prob. 34E Ch. 2.1 - Prob. 35E Ch. 2.1 - Prob. 36E Ch. 2.1 - Prob. 37E Ch. 2.1 - Prob. 38E Ch. 2.1 - Prob. 39E Ch. 2.1 - Prob. 40E Ch. 2.1 - Prob. 41E Ch. 2.1 - Prob. 42E Ch. 2.1 - Prob. 43E Ch. 2.1 - Prob. 44E Ch. 2.1 - Prob. 45E Ch. 2.1 - Prob. 46E Ch. 2.1 - Prob. 47E Ch. 2.1 - Prob. 48E Ch. 2.1 - Prob. 49E Ch. 2.1 - Prob. 50E Ch. 2.1 - 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