**Calculus: Graphical, Numerical, Algebraic, 3rd Edition Answers Ch 1 Prerequisites for Calculus Ex 1.1**

Calculus: Graphical, Numerical, Algebraic Answers

**Chapter 1 Prerequisites for Calculus Exercise 1.1 1E**

The given coordinates are A(1, 2) and B(-1, -1)

Now to find the increments in coordinate we subtract one coordinate point from the other, as shown below:

The increment in x-coordinate is:

**Chapter 1 Prerequisites for Calculus Exercise 1.1 1QR**

**Chapter 1 Prerequisites for Calculus Exercise 1.1 2E**

The given coordinates are A(3, 2) and B(-1, -2)

Now to find the increments in coordinate we subtract one coordinate point from the other, as shown below:

The increment in x-coordinate is:

**Chapter 1 Prerequisites for Calculus Exercise 1.1 2QR**

**Chapter 1 Prerequisites for Calculus Exercise 1.1 3E**

The given coordinates are A(-3, 1) and B(-8, -1)

Now to find the increments in coordinate we subtract one coordinate point from the other, as shown below:

The increment in x-coordinate is:

**Chapter 1 Prerequisites for Calculus Exercise 1.1 3QR **

**Chapter 1 Prerequisites for Calculus Exercise 1.1 4E **

The given coordinates are A(0, 4) and B(0, -2)

Now to find the increments in coordinate we subtract one coordinate point from the other, as shown below:

The increment in x-coordinate is:

**Chapter 1 Prerequisites for Calculus Exercise 1.1 4QR **

**Chapter 1 Prerequisites for Calculus Exercise 1.1 5E **

**Chapter 1 Prerequisites for Calculus Exercise 1.1 5QR **

The given equation is 3(x) – 4(y) = 5

(a) Replacing the given ordered pair in the equation, as below,

**Chapter 1 Prerequisites for Calculus Exercise 1.1 6E **

**Chapter 1 Prerequisites for Calculus Exercise 1.1 6QR **

The given equation is y = -2x + 5

**Chapter 1 Prerequisites for Calculus Exercise 1.1 7E **

**Chapter 1 Prerequisites for Calculus Exercise 1.1 7QR **

The given points are (1, 0) and (0, 1)

The distance between two points is given as

**Chapter 1 Prerequisites for Calculus Exercise 1.1 8E **

**Chapter 1 Prerequisites for Calculus Exercise 1.1 8QR **

**Chapter 1 Prerequisites for Calculus Exercise 1.1 9E **

The given point is P(3, 2)

(a) As we know that tor a vertical line we nave m = ∞ (undefined)

In case ot a vertical line we can see that the x-coordinate does not change but remains constant.

Therefore, the required equation is, x = 3

**Chapter 1 Prerequisites for Calculus Exercise 1.1 9QR **

The given equation is

4x – 3y = 7

Now as required in the question, the value ot y in terms ot x can be calculated as below:

**Chapter 1 Prerequisites for Calculus Exercise 1.1 10E **

**Chapter 1 Prerequisites for Calculus Exercise 1.1 10QR**

The given equation is

—2x + 5y = —3

Now as required in the question, the value of y in terms of x can be calculated as below:

**Chapter 1 Prerequisites for Calculus Exercise 1.1 11E**

The given point is P(0, -√2)

(a) As we know that tor a vertical line we have m = ∞(undefined)

In case ot a vertical line we can see that the x-coordinate does not change but remains constant,

Therefore, the required equation is, x = 0

This equation is also the equation tor the y-axis

**Chapter 1 Prerequisites for Calculus Exercise 1.1 12E**

The given point is P(-π, 0)

(a) As we know that tor a vertical line we nave m = ∞(undefined)

In case ot a vertical line we can see that the x-coordinate does not change but remains constant,

Therefore, the required equation is x = -π,

**Chapter 1 Prerequisites for Calculus Exercise 1.1 13E**

The given point is P(1, 1) and slope is m = 1

**Chapter 1 Prerequisites for Calculus Exercise 1.1 14E**

The given point is P(-1, 1) and slope is m = -1

**Chapter 1 Prerequisites for Calculus Exercise 1.1 15E**

The given point is P(0, 3) and slope is m = 2

**Chapter 1 Prerequisites for Calculus Exercise 1.1 16E**

The given point is P(-4, 0) and slope is m = -2

**Chapter 1 Prerequisites for Calculus Exercise 1.1 17E**

The given point is slope is m = 3 and the intercept b = -2

As we know that the slope-intercept torm ot equation is given by

y = mx + b

Where,

m is the slope ot the line,

b is the given intercept ot the line,

**Chapter 1 Prerequisites for Calculus Exercise 1.1 18E**

The given point is slope is m = -1 and the intercept b = 2

As we know that the slope-intercept torm ot equation is given by

y = mx + b

Where,

m is the slope ot the line,

b is the given intercept ot the line,

**Chapter 1 Prerequisites for Calculus Exercise 1.1 19E**

The given point is slope is m = -1/2 and the intercept b = -3

As we know that the slope-intercept torm ot equation is given by

y = mx + b

Where,

m is the slope ot the line,

b is the given intercept ot the line,

**Chapter 1 Prerequisites for Calculus Exercise 1.1 20E**

The given point is slope is m = 1/3 and the intercept b = -1

As we know that the slope-intercept torm ot equation is given by

y = mx + b

Where,

m is the slope ot the line,

b is the given intercept ot the line,

**Chapter 1 Prerequisites for Calculus Exercise 1.1 21E**

The given points on the required line are (0, 0) and (2, 3)

We know that the general torm ot the line is given as,

Ax+By=C

Where A and B are nonzero terms.

**Chapter 1 Prerequisites for Calculus Exercise 1.1 22E**

The given points on the required line are (1, 1) and (2, 1)

We know that the general torm ot the line is given as,

Ax+By=C

Where A and B are nonzero terms.

**Chapter 1 Prerequisites for Calculus Exercise 1.1 23E**

The given points on the required line are (-2, 0) and (-2, -2)

We know that the general torm ot the line is given as,

Ax+By=C

Where A and B are nonzero terms.

**Chapter 1 Prerequisites for Calculus Exercise 1.1 24E**

The given points on the required line are (-2, 0) and (-2, -2)

We know that the general torm ot the line is given as,

Ax+By=C

Where A and B are nonzero terms.

**Chapter 1 Prerequisites for Calculus Exercise 1.1 25E**

As given in the question, trom the grapn we see that the line contains the two points are (0, 0) and (10, 25)

**Chapter 1 Prerequisites for Calculus Exercise 1.1 26E**

As given in the question, trom the grapn we see that the line contains the two points are (0, 0) and (5, 2)

**Chapter 1 Prerequisites for Calculus Exercise 1.1 27E**

(b) On comparing the two equations we get

The y-intercept of the line is b = 3

(c) The graphical representation ot the line will be as shown below

[-10,10] by [-10,10]

**Chapter 1 Prerequisites for Calculus Exercise 1.1 28E**

The given equation of the line is,

x+y=2

This equation can be restructured as,

y=-x+2

Therefore this equation can be compared witn the slope-intercept for of the equation given as

y = mx + b

(a) On comparing the two equations we get

The slope of the line is m = -1

(b) On comparing the two equations we get

The y-intercept of the line is b = 2

(c) The graphical representation ot the line will be as shown below

[-10,10] by [-10,10]

**Chapter 1 Prerequisites for Calculus Exercise 1.1 29E**

(b) On comparing the two equations we get

The y-intercept of the line is b = 4

(c) The graphical representation ot the line will be as shown below

[-10,10] by [-10,10]

**Chapter 1 Prerequisites for Calculus Exercise 1.1 30E**

The given equation of the line is,

y=2x+4

Therefore this equation can be compared witn the slope-intercept for of the equation given as

y = mx + b

(a) On comparing the two equations we get

The slope of the line is m = 2

(b) On comparing the two equations we get

The y-intercept of the line is b = 2

(c) The graphical representation ot the line will be as shown below

[-10,10] by [-10,10]

**Chapter 1 Prerequisites for Calculus Exercise 1.1 31E**

The given equation of the line is,

y=-x+2

And the given point is P(0, 0)

Therefore this equation can be compared witn the slope-intercept for of the equation given as

y = mx + b

On comparing the two equations we get

The slope ottne line is m = -1

**Chapter 1 Prerequisites for Calculus Exercise 1.1 32E**

The given equation of the line is,

2x+x=4

And the given point is P(-2, 2)

Therefore this equation can be compared witn the slope-intercept for of the equation given as

y = mx + b

On comparing the two equations we get

The slope ottne line is m = -2

**Chapter 1 Prerequisites for Calculus Exercise 1.1 33E**

The given equation of the line is,

x=5

And the given point is P(-2, 4)

Therefore this equation can be compared witn the slope-intercept for of the equation given as

y = mx + b

On comparing the two equations we get

The slope ottne line is undefined as the line is vertical.

**Chapter 1 Prerequisites for Calculus Exercise 1.1 34E**

The given equation of the line is,

y=3

And the given point is P(-1, 1/2)

Therefore this equation can be compared witn the slope-intercept for of the equation given as

y = mx + b

On comparing the two equations we get

The slope ottne line is m=0

**Chapter 1 Prerequisites for Calculus Exercise 1.1 35E**

The given function is

f(x) = mx + b

**Chapter 1 Prerequisites for Calculus Exercise 1.1 36E**

The given function is

f(x) = mx + b

**Chapter 1 Prerequisites for Calculus Exercise 1.1 37E**

**Chapter 1 Prerequisites for Calculus Exercise 1.1 38E**

**Chapter 1 Prerequisites for Calculus Exercise 1.1 39E**

As asked in the question, we have to write the slope intercept form of the equation for the line passing through the points (—2, —1) and (3, 4)

**Chapter 1 Prerequisites for Calculus Exercise 1.1 40E**

**Chapter 1 Prerequisites for Calculus Exercise 1.1 41E**

**Chapter 1 Prerequisites for Calculus Exercise 1.1 42E**

(d) As the best insulator will have the largest temperature change per inch as this will allow larger temperature changes on the other side of thin walls. Therefore, the best insulator is fiberglass insulation whereas the poorest is gypsum wallboard.

**Chapter 1 Prerequisites for Calculus Exercise 1.1 43E**

**Chapter 1 Prerequisites for Calculus Exercise 1.1 44E**

**Chapter 1 Prerequisites for Calculus Exercise 1.1 45E**

**Chapter 1 Prerequisites for Calculus Exercise 1.1 46E**

**Chapter 1 Prerequisites for Calculus Exercise 1.1 47E**

**Chapter 1 Prerequisites for Calculus Exercise 1.1 48E**

**Chapter 1 Prerequisites for Calculus Exercise 1.1 49E**

**Chapter 1 Prerequisites for Calculus Exercise 1.1 50E**

**Chapter 1 Prerequisites for Calculus Exercise 1.1 51E**

**Chapter 1 Prerequisites for Calculus Exercise 1.1 52E**

**Chapter 1 Prerequisites for Calculus Exercise 1.1 53E**

**Chapter 1 Prerequisites for Calculus Exercise 1.1 54E**

**Chapter 1 Prerequisites for Calculus Exercise 1.1 55E**

The three points can be joined in three different ways to torm three different parallelograms as shown below:

From the above three graphs we can see that the three missing pints are (5, 2), (-1, 4), and (-1, -2) respectively.

**Chapter 1 Prerequisites for Calculus Exercise 1.1 56E**

We nave to snow that it the midpoints ot consecutive sides ot any quadrilateral are connected, then the resulting figure is a parallelogram.

From the above, the slopes of the four lines, we can see that two slopes each are of equal values. This means that those two lines are parallel to each other.

Therefore from here we conclude that if the midpoints of consecutive sides of any quadrilateral are connected, then the resulting figure is a parallelogram.

**Chapter 1 Prerequisites for Calculus Exercise 1.1 57E**

As given in the question, the radius of the circle is 5, and the centre of the circle is at (0, 0) . The tangent passes through the point (3, 4).

**Chapter 1 Prerequisites for Calculus Exercise 1.1 58E**