
<Parag>Here you can find calculus lessons that will help you understand by explaining in detail and clarifying with some examples.</Parag>


Here you can find calculus lessons that will help you understand by explaining in detail and clarifying with some examples.

Calculus

# Plotting Points on a Coordinate Plane (Coordinates on a Graph)
<Parag>The Cartesian Coordinate System is two perpendicular and graduated lines that intersect at a point <Tex>O</Tex> called **the origin of the coordinate axis**.</Parag>
<Parag>In the plane attributed to this system of coordinates, we can mark any point <Tex>M(x;y)</Tex> with a pair of coordinates(<Tex>x</Tex>-coordinate and <Tex>y</Tex>-coordinate).
Before we get to know <Tex>x</Tex> and <Tex>y</Tex>, let's first get to know the graphic of the Coordinate Axis.</Parag>

<Image src="/imgs/lessons/calculus/coordinate-plane.png" alt="empty coordinate plan with x a,d y axis" width='360px' height='360px'/>

<Parag>Perpendicular lines make up four quadrants, as shown below.</Parag>

<Image src="/imgs/lessons/calculus/coordinate-plane-quadrants.png" alt="coordinate plan quadrants" width='360px' height='360px'/>

<Parag>The two graduated axes have their own name:</Parag>
<Parag>- The horizontal axis is called the **<Tex>x</Tex>-axis** or **the abscissa axis**, and is graded incremental from left to right.
- The vertical axis is called the **<Tex>y</Tex>-axis** or **the ordinates axis**, and it is graded incremental from bottom to top.</Parag>
<Parag>The point of their intersection is called **the origin <Tex>O</Tex>** of The Cartesian Coordinate System, and we write <Tex>O(0;0).</Tex></Parag>

## How to plotting a given point <Tex>M(x;y)</Tex> on a coordinate plane.
- We mark the real number <Tex>x</Tex> on the <Tex>x</Tex>-axis and drawing a line (1) pass through <Tex>x</Tex> and parallel to the <Tex>y</Tex>-axis.
- We mark the real number <Tex>y</Tex> on the <Tex>y</Tex>-axis and drawing a line (2) pass through <Tex>y</Tex> and parallel to the <Tex>x</Tex>-axis.
- The intersection of the two lines (1) and (2) is the coordinates of the point <Tex>M.</Tex>

<Image src="/imgs/lessons/calculus/coordinate-plane-m.png" alt="coordinate plan of a given point M" width='360px' height='360px'/>

**Example1:**
<Parag>Let's plot the point <Tex>A(2;-3)</Tex> on the coordinates axis.</Parag>
<Parag>We mark the number 2 in the <Tex>x</Tex>-axis, and we mark the number -3 in the <Tex>y</Tex>-axis.</Parag>
<Image src="/imgs/lessons/calculus/coordinate-plane-ex1.png" alt="coordinate plan example1" width='360px' height='360px'/>
<Parag>Let's plot the point <Tex>P(0;6)</Tex> on the coordinates axis.</Parag>
<Parag>We mark the number 0 in the <Tex>x</Tex>-axis, and we mark the number 6 in the <Tex>y</Tex>-axis.</Parag>
<Note>When an abscissa is zero <Tex>(x=0)</Tex>, the point belongs to the ordinate axis.</Note>
<Image src="/imgs/lessons/calculus/coordinate-plane-ex2.png" alt="coordinate plan example2" width='360px' height='360px'/>
<Parag>Let's plot the point <Tex>D(-3;0)</Tex> on the coordinates axis.</Parag>
<Parag>We mark the number -3 in the <Tex>x</Tex>-axis, and we mark the number 0 in the <Tex>y</Tex>-axis.</Parag>
<Note>When an ordinate is zero <Tex>(y=0)</Tex>, the point belongs to the abscissa axis.</Note>
<Image src="/imgs/lessons/calculus/coordinate-plane-ex3.png" alt="coordinate plan example3" width='360px' height='360px'/>
<Parag>Let's plot the point <Tex>Q(1;5)</Tex> on the coordinates axis.</Parag>
<Parag>We mark the number 1 in the <Tex>x</Tex>-axis, and we mark the number 5 in the <Tex>y</Tex>-axis.</Parag>
<Image src="/imgs/lessons/calculus/coordinate-plane-ex4.png" alt="coordinate plan example4" width='360px' height='360px'/>
<Parag>Let's plot the point <Tex>F(-6;-4)</Tex> on the coordinates axis.</Parag>
<Parag>We mark the number -6 in the <Tex>x</Tex>-axis, and we mark the number -4 in the <Tex>y</Tex>-axis.</Parag>
<Image src="/imgs/lessons/calculus/coordinate-plane-ex5.png" alt="coordinate plan example4" width='360px' height='360px'/>

Learn how to plotting points on acoordinate plane using the given coordinates of a point

Plotting Points on a Coordinate Plane

# Midpoint and Distance Formula on the Coordinate Plane

<Parag>In the Cartesian Coordinate System, we take two different ponts <Tex>A(x_1;y_1)</Tex> and <Tex>B(x_2;y_2)</Tex>. These two points form a **line segment** with endpoints <Tex>A</Tex> and <Tex>B</Tex>.</Parag>

<Parag>We know that each line segment has the following two properties:</Parag>
- Length (the distance between the endpoints)
- Midpoint (the point in the center of the line segment)

<Parag>In this lesson, we will see how to calculate the length of the segment and how to find the coordinates of the midpoint, which is very easy by applying the rules that we will see shortly.</Parag>

## Distance Formula

<Parag>To measure the distance between two points in an orthogonal coordinate system, we use a formula that is very important in geometry. This formula is actually extracted from the Pythagorean Theorem as it is written below:</Parag>

<Tex block>a^2+b^2=d^2</Tex>

<Parag>In which:</Parag>

- <Tex>d</Tex> is the length of the hypotenuse of a right triangle (the longest side).
- <Tex>a</Tex> and <Tex>b</Tex> the length of the two others sides (the shortest sides).

<Parag>As a reminder, the Pythagorean Theorem only applies to a right-angled triangle</Parag>

<Image src="/imgs/lessons/calculus/pyth_theo.png" alt="pythagorean theorem formula" width='386px' height='355px'/>

<Note>
Let <Tex>A(x_1;y_1)</Tex> and <Tex>B(x_2;y_2)</Tex> two points. The distance between <Tex>A</Tex> and <Tex>B</Tex> is denoted by <Tex>d</Tex>, and we have:

<Tex block>d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}</Tex>

</Note>

<Parag>But how did we get this formula? Let's first look at the figure below.</Parag>

<Image src="/imgs/lessons/calculus/distance_formula_proof.png" alt="distance formula in coordinate plane system" width='511px' height='304px'/>

- The number <Tex>d</Tex> is the length of the hypotenuse (Distance between <Tex>A</Tex> and <Tex>B</Tex>)
- The number <Tex>a=x_2-x_1</Tex> is the length of the horizontal leg.
- The number <Tex>b=y_2-y_1</Tex> is the length of the vertical leg.

<Parag>So, applying the Pythagorean theorem, we have:</Parag>

<Tex block>
\begin{align*}
d^2 &= a^2+b^2 \\\
d^2 &= (x_2-x_1)^2+(y_2-y_1)^2 \\\
d &= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} 
\end{align*}
</Tex>

<Parag>Remember that the distance is always positive.</Parag>

## Midpoint Formula

<Parag>Using geometry tools (for example, Compass), we can easily determine the midpoint of a line segment. But isn't there another way, without using geometry tools, that we can determine the midpoint?</Parag>

<Parag>Yes, of course, there is an arithmetic method that depends on knowing the coordinates of the two endpoints of the line segment in an orthogonal coordinate system, using a formula to calculate the coordinates of the center.</Parag>

<Parag>We take two points <Tex>A(x_1;y_1)</Tex> and <Tex>B(x_2;y_2)</Tex> in an orthogonal coordinate system, and take the point <Tex>M</Tex> as the midpoint of the line segment <Tex>\overline{AB}</Tex>.</Parag>

<Note>
We say that <Tex>M</Tex> is the midpoint of the line segment <Tex>\overline{AB}</Tex> if the following two conditions are met together:

- There is the same distance between the two endpoints of the line segment, i.e. <Tex>\overline{MA}=\overline{MB}</Tex>
- The point <Tex>M</Tex> belongs to the line segment <Tex>\overline{AB}</Tex>.
</Note>

<Parag>To calculate the midpoint coordinates, we use the following property:</Parag>

<Note>

Let <Tex>A(x_1;y_1)</Tex> and <Tex>B(x_2;y_2)</Tex> two points and <Tex>M</Tex> the midpoint of the line segment <Tex>\overline{AB}</Tex>. We have:

<Tex block>M\left(\frac{x_1+x_2}{2};\frac{y_1+y_2}{2}\right)</Tex>

</Note>

<Parag>The following figure shows the coordinates of the midpoint of a segment in a coordinate plane.</Parag>

<Image src="/imgs/lessons/calculus/midpoint_formula.png" alt="midpoint formula in coordinate plane system" width='407px' height='350px'/>

<Parag>Let's take some examples to calculate distance and midpoint coordinates.</Parag>

---------

**Example1:**

<Parag>Consider the points <Tex>A(2;5)</Tex> and <Tex>B(-3;1)</Tex>.</Parag>

1. Calculate <Tex>d</Tex> the distance of <Tex>\overline{AB}</Tex>.
2. Calculate the coordinates of <Tex>M</Tex> the midpoint of the segment <Tex>\overline{AB}</Tex>.
3. Deduce the lengths of the two segments <Tex>\overline{MA}</Tex> and <Tex>\overline{MB}</Tex>.

#### Solution:

<Parag>1) Plotting the two points <Tex>A</Tex> and <Tex>B</Tex> in the coordinate plane.</Parag>

<Image src="/imgs/lessons/calculus/ex1_1.png" alt="distance between A and B" width='544px' height='427px'/>

<Parag>Let <Tex>(x_1;y_1)=(2;5)</Tex> and <Tex>(x_2;y_2)=(-3;1)</Tex>; then we substitute in the distance formula:</Parag>

<Tex block>
\begin{align*}
d &= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\\
&= \sqrt{(-3-2)^2+(1-5)^2} \\\
&= \sqrt{(-5)^2+(-4)^2} \\\
&= \sqrt{25+16} \\\
&= \sqrt{41}
\end{align*}
</Tex>

<Parag>2) We calculate the coordinates of <Tex>M</Tex> using the midpoint formula:</Parag>

<Tex block>
M\left(\frac{x_1+x_2}{2};\frac{y_1+y_2}{2}\right) \\\
M\left(\frac{2+(-3)}{2};\frac{5+1}{2}\right) \\\
M\left(\frac{-1}{2};\frac{6}{2}\right) \\\
M\left(\frac{-1}{2};3\right)
</Tex>

<Image src="/imgs/lessons/calculus/ex1_2.png" alt="midpoint of the line segment AB" width='432px' height='430px'/>

<Parag>3) <Tex>M</Tex> is the midpoint of the line segment <Tex>\overline{AB}</Tex>, This means that there is the same distance from both endpoints of the segment</Parag>

<Parag>Therefore, <Tex>\overline{MA}=\overline{MB}=\frac{\overline{AB}}{2}=\frac{\sqrt{41}}{2}</Tex></Parag>

**Example2:**

<Parag>Consider the points <Tex>C(2;1)</Tex> and <Tex>D(6;3)</Tex>.</Parag>

1. Calculate <Tex>d</Tex> the distance of <Tex>\overline{CD}</Tex>.
2. Calculate the coordinates of <Tex>P</Tex> the midpoint of the segment <Tex>\overline{CD}</Tex>.
3. Deduce the lengths of the two segments <Tex>\overline{PC}</Tex> and <Tex>\overline{PD}</Tex>.

#### Solution:

<Parag>1) Pltting the two points <Tex>C</Tex> and <Tex>D</Tex> in the coordinate plane</Parag>.

<Image src="/imgs/lessons/calculus/ex2_1.png" alt="midpoint of the line segment AB" width='490px' height='312px'/>

<Parag>Let <Tex>(x_1;y_1)=(2;1)</Tex> and <Tex>(x_2;y_2)=(6;3)</Tex>; then we substitute in the distance formula:</Parag>

<Tex block>
\begin{align*}
d &= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\\
&= \sqrt{(6-2)^2+(3-1)^2} \\\
&= \sqrt{(4)^2+(2)^2} \\\
&= \sqrt{16+4} \\\
&= \sqrt{20} \\\
&= \sqrt{4} \times \sqrt{5} \\\
&= 2\sqrt{5}
\end{align*}
</Tex>

<Parag>2) We calculate the coordinates of <Tex>P</Tex> using the midpoint formula:</Parag>

<Tex block>
P\left(\frac{x_1+x_2}{2};\frac{y_1+y_2}{2}\right) \\\
P\left(\frac{2+6}{2};\frac{3+1}{2}\right) \\\
P\left(\frac{8}{2};\frac{4}{2}\right) \\\
P\left(4;2\right)
</Tex>

<Image src="/imgs/lessons/calculus/ex2_2.png" alt="midpoint of the line segment AB" width='445px' height='271px'/>

<Parag>3) <Tex>P</Tex> is the midpoint of the line segment <Tex>\overline{CD}</Tex>, This means that there is the same distance from both endpoints of the segment.</Parag>

<Parag>Therefore, <Tex>\overline{PC}=\overline{PD}=\frac{\overline{CD}}{2}=\frac{2\sqrt{5}}{2}=\sqrt{5}</Tex></Parag>

**Example3:**

<Parag>Consider the points <Tex>E(3;4)</Tex> and <Tex>F(-7;-1)</Tex>.</Parag>

1. Calculate the distance between <Tex>E</Tex> and the origin.
2. Calculate the distance between <Tex>F</Tex> and the origin.

#### Solution:

<Parag>1) The origin of the coordinate plane is the point <Tex>O(0;0)</Tex>.</Parag>

<Image src="/imgs/lessons/calculus/ex3_1.png" alt="midpoint of the line segment AB" width='399px' height='389px'/>

<Parag>Let <Tex>(x_1;y_1)=(0;0)</Tex> and <Tex>(x_2;y_2)=(3;4)</Tex>; then we substitute in the distance formula:</Parag>

<Tex block>
\begin{align*}
d &= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\\
&= \sqrt{(3-0)^2+(4-0)^2} \\\
&= \sqrt{(3)^2+(4)^2} \\\
&= \sqrt{9+16} \\\
&= \sqrt{25} \\\
&= 5
\end{align*}
</Tex>

<Parag>2) In the same way we calculate the distance between <Tex>F</Tex> and the origin.</Parag>

<Image src="/imgs/lessons/calculus/ex3_2.png" alt="midpoint of the line segment AB" width='554px' height='247px'/>

<Parag>Let <Tex>(x_1;y_1)=(0;0)</Tex> and <Tex>(x_2;y_2)=(-7;-1)</Tex>; then we substitute in the distance formula:</Parag>

<Tex block>
\begin{align*}
d &= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\\
&= \sqrt{(-7-0)^2+(-1-0)^2} \\\
&= \sqrt{(-7)^2+(-1)^2} \\\
&= \sqrt{49+1} \\\
&= \sqrt{50} \\\
&= \sqrt{25} \times \sqrt{2} \\\
&= 5\sqrt{2} 
\end{align*}
</Tex>

**Example4:**

<Parag>Consider the points <Tex>A(-1;1)</Tex> and <Tex>M(1;2)</Tex>.</Parag>

1. Determine the coordinates of <Tex>B</Tex> if <Tex>M</Tex> is the midpoint of the line segment <Tex>\overline{AB}</Tex>
2. Calculate the distance of the line segment <Tex>\overline{AB}</Tex>

#### Solution:

<Parag>1) Unlike examples 1 and 2, in this example we have the coordinates of the midpoint but we don't know the coordinates of the endpoint <Tex>B</Tex> of the line segment <Tex>\overline{AB}</Tex>. To solve this problem, it's very easy, we just apply the Midpoint Formula but this time differently.</Parag>

<Parag>We have <Tex>A(-1;1)</Tex> and let <Tex>B(x_2;y_2)</Tex></Parag>

<Image src="/imgs/lessons/calculus/ex4_1.png" alt="midpoint of the line segment AB" width='448px' height='320px'/>

<Parag>Since <Tex>M</Tex> is the midpoint of <Tex>\overline{AB}</Tex> then:</Parag>

<Tex block>M\left(\frac{-1+x_2}{2};\frac{1+y_2}{2}\right)=M(1;2)</Tex>

<Parag>The coordinates on both sides of the equation are equal. It means that the <Tex>x</Tex>-coordinate on the left side is equal to the <Tex>x</Tex>-coordinate on the right side.</Parag>

<Parag>The mathematical expression is:</Parag>

<Tex block>\frac{-1+x_2}{2}=1 \tag{1}</Tex>

<Parag>And the <Tex>y</Tex>-coordinate on the left side is equal to the <Tex>y</Tex>-coordinate on the right side.</Parag>

<Parag>The mathematical expression is:</Parag>

<Tex block>\frac{1+y_2}{2}=2 \tag{2}</Tex>

<Parag>Here we get two simple equations:</Parag>

- The unknown in equation (1) is <Tex>x_2</Tex>
- The unknown in equation (2) is <Tex>y_2</Tex>

<Parag>Let's solve equation (1), we have</Parag>

<Tex block>
\begin{align*}
\frac{-1+x_2}{2} &=1 \\\
-1+x_2 &=1\times2 \\\
-1+x_2 &=2 \\\
x_2 &=2+1 \\\
x_2 &=3
\end{align*}
</Tex>

<Parag>Let's solve equation (2), we have:</Parag>

<Tex block>
\begin{align*}
\frac{1+y_2}{2} &=2 \\\
1+y_2 &=2\times2 \\\
1+y_2 &=4 \\\
y_2 &=4-1 \\\
y_2 &=3
\end{align*}
</Tex>

<Parag>Finally, the coordinates of the endpoint <Tex>B</Tex> of the line segment <Tex>\overline{AB}</Tex> are <Tex>B(3;3)</Tex></Parag>

<Image src="/imgs/lessons/calculus/ex4_2.png" alt="midpoint of the line segment AB" width='445px' height='313px'/>

<Parag>2) Let's calculate the distance <Tex>d</Tex> of the line segment<Tex>\overline{AB}</Tex></Parag>

<Parag>Let <Tex>(x_1;y_1)=(3;-4)</Tex> and <Tex>(x_2;y_2)=(3;3)</Tex></Parag>

<Parag>Therefore, the length of the line segment <Tex>\overline{AB}</Tex> is:</Parag>

<Tex block>
\begin{align*}
d &= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\\
&= \sqrt{(3-3)^2+(3-(-4))^2} \\\
&= \sqrt{0^2+7^2} \\\
&= \sqrt{0+49} \\\
&= \sqrt{49} \\\
&= 7
\end{align*}
</Tex>

In this lesson you will learn how to find the midpoint of a Line Segment and The distance between two points using Midpoint Formula and Distance Formula

Midpoint and Distance Formula


# Constant function, Linear function and Affine function and their graphic representation

The functions that we will see in this lesson are very simple functions with only one variable, but they are essential to understanding the concept of numeric functions later.

## Constant Function

<Note>

A Constant Function is every function written in the form <Tex>f(x)=b</Tex> where <Tex>b</Tex> is a known real number.

</Note>

From the formula for the constant function, we can see that the value of <Tex>f(x)</Tex> is completely independent of the value of <Tex>x</Tex>, in other words, the value of <Tex>f(x)</Tex> does not change if we change the value of <Tex>x</Tex>.

As an example from reality, if we have a car traveling at a constant speed over a period of time (the variable is time t), then the expression of the function in this case is <Tex>f(t)=v</Tex> in which <Tex>v</Tex> is the velocity of the car.

#### Let represent graphically the two defined functions <Tex>f(x)=3</Tex> and <Tex>g(x)=-2</Tex> as follows:

<Image src="/imgs/lessons/calculus/constant-function.png" alt="represent graphically a constant function" width='426px' height='481px'/>

- The number 3 is the intersection of the straight line that passes through the ordinate 3 (<Tex>y</Tex>-axis) and is parallel to the axis of the abscissa (<Tex>x</Tex>-axis).

- The number -2 is the intersection of the straight line that passes through the ordinate -2 (<Tex>y</Tex>-axis) and is parallel to the axis of the abscissa (<Tex>x</Tex>-axis).

## Linear Function

A Linear Function is any function written in the form <Tex>f(x)=ax</Tex> where <Tex>a</Tex> is a known real number and is called **the slope of the line**.

Through the formula of the linear function, we can see that the value of <Tex>f(x)</Tex> changes with the change of <Tex>x</Tex> as follows:

<Tex block>
f(1)=a\times1\\\
f(2)=a\times2\\\
f(3)=a\times3\\\
f(-1)=a\times(-1)\\\
</Tex>

The graphic representation of a linear function is a line that passes through the coordinates origin <Tex>O(0;0)</Tex>.

So how do we represent the line of a linear function in an orthonormal coordinate system?
We said earlier that the graphic representation of a linear function is a straight line. So we need two different points to create this line. How will we get them?

- The first point is <Tex>O(0;0)</Tex> because the line passes through the origin of the coordinate plane.

- The second point we get by substituting <Tex>x</Tex> into the <Tex>(x;f(x))</Tex> coordinates.

That is, if we take <Tex>x=2</Tex>, then:<Tex>f(2)=2a.</Tex>

**Example1:**

Let us represent graphically the linear function <Tex>f(x)=3x</Tex>.

The line passes from <Tex>O(0;0)</Tex> and another point <Tex>A(x;f(x))</Tex>

Let's take for example <Tex>x=1</Tex>, so <Tex>f(1)=3\times1=3</Tex>.

We get the second point:

<Text block>A(1;f(1))=A(1;3)</Tex>

Let us represent graphically the linear function <Tex>g(x)=-2x</Tex>.

The line passes from <Tex>O(0;0)</Tex> and another point <Tex>B(x;g(x))</Tex>

Let's take for example <Tex>x=2</Tex>, so <Tex>g(2)=-2\times4=-4</Tex>.

We get the second point:

<Text block>B(1;g(2))=B(1;-4)</Tex>

<Image src="/imgs/lessons/calculus/linear-function.png" alt="represent graphically a linear function" width='462px' height='617px'/>

## Affine Function

An Affine Function is any function written in the form <Tex>f(x)=ax+b</Tex> where <Tex>a</Tex> is a known real number and is called **the slope of the line**, and b is known real number called **the y-intercept** with the line.

Through the formula of the affine function, we can see that the value of <Tex>f(x)</Tex> changes with the change of <Tex>x</Tex> as follows:

<Tex block>
f(1)=a\times1+b\\\
f(2)=a\times2+b\\\
f(3)=a\times3+b\\\
f(-1)=a\times(-1)+b\\\
</Tex>

The graphic representation of the affine function intersects with the <Tex>y</Tex>-axis. So how do we represent the line of an affine function in an orthonormal coordinate system?

The method is not very different from the linear function. First we represent <Tex>b</Tex> on the <Tex>y</Tex>-axis as the first point.

As for the second point, we will get it by substituting <Tex>x</Tex> into the formula for the affine function as follows: <Tex>(x; f(x))</Tex>

That is, if we take <Tex>x=2</Tex>, then:<Tex>f(2)=2a+b.</Tex>

**Example2:**

Let us represent graphically the affine function <Tex>f(x)=2x+1</Tex>.

We have <Tex>b=1</Tex>, so, the first point is <Tex>A(0;1)</Tex>.

<Image src="/imgs/lessons/calculus/affine-function-1-step1.png" alt="represent graphically an affine function" width='377px' height='364px'/>

Let's take for example <Tex>x=2</Tex>, therefore <Tex>f(2)=2\times2+1=5</Tex>.

Hence, the second point is <Tex>B(2;f(2))=B(2;5)</Tex>

<Image src="/imgs/lessons/calculus/affine-function-1-step2.png" alt="represent graphically an affine function" width='531px' height='579px'/>

**Example3:**

Let us represent graphically the affine function <Tex>f(x)=-x+2</Tex>.

We have <Tex>b=2</Tex>, so, the first point is <Tex>A(0;2)</Tex>.

<Image src="/imgs/lessons/calculus/affine-function-2-step1.png" alt="represent graphically an affine function" width='428px' height='403px'/>

Let's take for example <Tex>x=1</Tex>, therefore <Tex>f(1)=-1+2=1</Tex>.

Hence, the second point is <Tex>B(1;f(1))=B(1;1)</Tex>

<Image src="/imgs/lessons/calculus/affine-function-2-step2.png" alt="represent graphically an affine function" width='718px' height='383px'/>

Learn the most basic types of functions, Constant, Linear and Affine Functions and their graphic representation in the coordinate plane.

Constant function, Linear function and Affine function


# What is the Slope of a Line

**The slope** or **Rate of Change** of a line is a measurement of the rate at which it rises or falls. A rising line has a positive slope whereas a falling line has a negative slope. More specifically, slope is a number that is a ratio that describes the steepness of a line. The larger the absolute value of the slope, the steeper the line and vice versa.

<Image src="/imgs/lessons/calculus/rise_run.jpeg" alt="rise and run of the slope of a line" width={500} height={319} placeholder="blur" blurDataURL=""/>

**RISE** is how much a line goes up or down.

**RUN** is how much a line moves left or right.

<Tex block>Slope=\dfrac{RISE}{RUN}</Tex>

**Example** A line with a slop of <Tex>\dfrac{2}{1}</Tex>

A slope of <Tex>\dfrac{2}{1}</Tex> means that every time the line rises 2, it also runs 1.

<Image src="/imgs/lessons/calculus/slope-rise-run-example.png" alt="rise and run of the slope of a line example" width={236} height={300}/>

## Types of Slope of a Stright Line

Generally, there are four types of slope of a line, namely positive, negative, zero slopes and undefined slope. Let's see in detail each type separately:

**Positive Slope:** rises (increasing) from left to right.

<Image src="/imgs/lessons/calculus/positive_slope.jpeg" alt="positive slope in graph" width={400} height={400} placeholder="blur" blurDataURL=""/>

**Negative Slope:** falls (decreasing) from left to right.

<Image src="/imgs/lessons/calculus/negative_slope.jpeg" alt="negative slope in graph" width={400} height={400}/>

**Zero Slope:** is horizontal because the rise is 0, and 0 divided by any number is 0.

<Image src="/imgs/lessons/calculus/zero_slope.jpeg" alt="zero slope in graph" width={400} height={400}/>

**Undefined Slope:** is a vertical line because the run is 0, and any number divided by 0 is undefined.

<Image src="/imgs/lessons/calculus/undefined_slope.jpeg" alt="undefined slope in graph" width={400} height={400} placeholder="blur" blurDataURL="" />

## 5 Things You Need to Know About Slope

1. Any time you move **UP**, that is a **POSITIVE RISE**.

2. Any time you move **DOWN**, that is a **NEGATIVE RISE**.

3. Any time you move **RIGHT**, that is a **POSITIVE RUN**.

4. Any time you move **LEFT**, that is a **NEGATIVE RUN**.

5. The slope is the **SAME** everywhere on a **STRAIGHT LINE**.

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<Example type="example" num="1"/>

In the graph below, the line rises 3 and runs 2.

<Image src="/imgs/lessons/calculus/slope_rise_up_example.jpeg" alt="undefined slope in graph" width={419} height={437} placeholder="blur" blurDataURL="" />

<Example type="example" num="2"/>

In the graph below, the line rises -2 and runs 3.

<Image src="/imgs/lessons/calculus/slope_rise_down_example.jpeg" alt="undefined slope in graph" width={444} height={325} placeholder="blur" blurDataURL="" />

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<Related>

<Link to="../calculus/constant-linear-affine-functions" as="/lessons/calculus/constant-linear-affine-functions">Constant function, Linear function and Affine function and their graphic representation</Link>

</Related>

Learn what is the slope (rate of change) of a line, and what is the types of slope of a line, this lesson is illustrated by graphs for each type of slope.

What is the Slope of a Line


# Finding a Slope of a Line Using 4 Methods

In this lesson, you will learn how to find a slope of a line using four different methods. But before that, perhaps here you will need to remember what we saw in Lesson <Link to="../calculus/what-is-the-slope-of-a-line" as="/lessons/calculus/what-is-the-slope-of-a-line">"What is the Slope of a Line"</Link>.

The slope of a line is the *steepness* of a line, or how quickly it rises (or falls) as we move from left to right. To measure this steepness, we define *run* to be the distance we move to the right, and *rise* to be the corresponding distance that the lines rises (or falls). The slope of a line is the **ratio** of rise and run:

<Tex block>\boxed{slope=\dfrac{RISE}{RUN}}</Tex>

The four ways to determine the slope of a line:

## Find the slope of the line from a table

To determine the slope of a line from a table.

**Step 1:** Identify the vertical change in each consecutive pair of *y-values* in the table. <Tex>vertical\\;change=y_{n+1}-y_n</Tex>

**Step 2:** Identify the horizontal change in each consecutive pair of *x-values* in the table. <Tex>horizontal\\;change=x_{n+1}-x_n</Tex>

**Step 3:** Write ratios showing the corresponding <Tex>\dfrac{Vchange}{Hchange}=\dfrac{y\_\{n+1}-y\_\{n}}{x\_\{n+1}-x\_\{n}}</Tex> in simplest form.

<Example type="example" num="1"/>

Fin the slope of line from the table bellow:

<Image src="/imgs/lessons/calculus/table_slope.png" alt="slope of line from table example" width={242} height={213} placeholder="blur" blurDataURL="" />

**Step 1:** The change in each consecutive pair of y-values:

<Image src="/imgs/lessons/calculus/slope_table_ychange.jpeg" alt="slope of line from table example" width={444} height={238} placeholder="blur" blurDataURL="" />

**Step 2:** The change in each consecutive pair of x-values:

<Image src="/imgs/lessons/calculus/slope_table_xchange.jpeg" alt="slope of line from table example" width={500} height={220} placeholder="blur" blurDataURL="" />

**Step 3:** Write the ratios: <Tex>\dfrac{4}{-2}=\dfrac{8}{-4}=\dfrac{6}{-3}=-2</Tex>.

The slope of the line is <Tex>-2</Tex>

--------------

## Find the slope of the line from a graph

To ind the slope of line from a graph:

**Step 1:** Pick any two points on the line. Get the coordinates of the two point <Tex>A(x_1;y_1)</Tex> and <Tex>B(x_2;y_2)</Tex>.

**Step 2:** Calculate the slope using the slope formula: <Tex>m=\dfrac{y_2-y_1}{x_2-x_1}</Tex>

<Image src="/imgs/lessons/calculus/slope_graph.jpeg" alt="slope of line from a graph example" width={500} height={220} placeholder="blur" blurDataURL="" />

<Example type="example" num="2"/>

Find the slope of a line in the graph below:

<Image src="/imgs/lessons/calculus/slope_graph_example.jpeg" alt="slope of line from a graph example" width={500} height={220} placeholder="blur" blurDataURL="" />

**Step 1:** Pick two different points in the graph (It is better to choose points whose coordinates are integers)

<Image src="/imgs/lessons/calculus/slope_graph_example2.jpeg" alt="slope of line from a graph example" width={500} height={220} placeholder="blur" blurDataURL="" />

**Step 2:** In this case, <Tex>(x_1;y_1)=(5;3)</Tex> and <Tex>(x_2;y_2)=(0;1)</Tex>, Thus, the slope is:

<Tex block>m=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{1-3}{0-5}=\dfrac{-4}{-5}=\dfrac{4}{5}</Tex>

----------

## Find the slope of line passing through two points

To find the slope of a line passing through two points, it is sufficient to use only the slope of the line formula.

If the line passes through <Tex>(x_1;y_1)</Tex> and <Tex>(x_2;y_2)</Tex>, therefor, the slope is:

<Tex block>m=\dfrac{y_2-y_1}{x_2-x_1}</Tex>

<Example type="example" num="3"/>

Find the slope of a line passing through the two points <Tex>A(1;2)</Tex> and <Tex>B(-1;5)</Tex>.

<Tex block>m=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{5-2}{-1-1}=\dfrac{3}{-2}=\dfrac{-3}{2}</Tex>

---------------

## Find the slope of the line from an equation

we use this method when we have the linear functions and the affine functions.

**Step 1:** Rewrite the equation in the form: <Tex>\boxed{y=mx+b}</Tex>.

**Step 2:** Slope is the rate of change, therefore, it's next to the variable <Tex>x</Tex>. i.e, the slope is <Tex>m</Tex> the coefficient of <Tex>x</Tex>.

<Example type="example" num="3"/>

Find the slope in each equation:

1.  <Tex>y=\dfrac{1}{2}x+4</Tex>

2.  <Tex>y=-3x-2</Tex>

3.  <Tex>y=\dfrac{3x}{2}-5</Tex>

4.  <Tex>y=-x+1</Tex>

5.  <Tex>y=x+\dfrac{1}{2}</Tex>

6.  <Tex>2(x+8)+3y=4</Tex>

<Example type="solution"/>

1. <Tex>y=\colorbox{aqua}{$\dfrac{1}{2}$}x+4</Tex>

Therefore, the slope is <Tex>m=\dfrac{1}{2}</Tex>

2. <Tex>y=\colorbox{aqua}{$-3$}x-2</Tex>

Therefore, the slope is <Tex>m=-3</Tex>

3. <Tex>y=\dfrac{3x}{2}-5=\colorbox{aqua}{$\dfrac{3}{2}$}x-5</Tex>

Therefore, the slope is <Tex>m=\dfrac{3}{2}</Tex>

4. <Tex>y=-x+1=\colorbox{aqua}{$-1$}x+1</Tex>

Therefore, the slope is <Tex>m=-1</Tex>

5. <Tex>y=x+\dfrac{1}{2}=\colorbox{aqua}{$1$}x+\dfrac{1}{2}</Tex>

Therefore, the slope is <Tex>m=1</Tex>

6. <Tex>2(x+8)+3y=4</Tex>

In this case, we should isolate y of the equation.

<Tex block>
\begin{align*}
2(x+8)+3y&=4\\\
2x+16+3y&=4\\\
3y&=-2x-16+4\\\
3y&=-2x-12\\\
y&=\dfrac{-2}{3}x-\dfrac{12}{3}\\\
y&=\dfrac{-2}{3}x-4\\\
\end{align*}
</Tex>

We got <Tex>y=\colorbox{aqua}{$\dfrac{-2}{3}$}x-4</Tex>

Therefore, the slope is <Tex>m=\dfrac{-2}{3}</Tex>

---------------

<Related>

<Link to="../calculus/what-is-the-slope-of-a-line" as="/lessons/calculus/what-is-the-slope-of-a-line">What is the Slope of a Line</Link><br/>
<Link to="../calculus/constant-linear-affine-functions" as="/lessons/calculus/constant-linear-affine-functions">Constant function, Linear function and Affine function and their graphic representation</Link>

</Related>

Learn how to find the slope of a line using 4 different methods:from a graph, from a table, from two points and from an equation of straight line.

Explorer all lessons of Calculus