Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 4 Writing Linear Functions Exercise 4.1

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 4 Writing Linear Functions Exercise 4.1

Page 96  Exercise 1  Answer

Given:

To find, find the slope and y-intercept

The slope of the line \(m=\frac{y_2-y_1}{x_2-x_1}\)

The slope-intercept form y=mx+b

Here​(x₁,y₁)=(2,3), (x₂,y₂)=(0,−1)
​
By substituting, we get

​m\(\frac{-1-3}{0-2}\)

\(=\frac{-4}{-2}\)

=2
​

In the slope-intercept form, we substitute m=2 and take any point we get
​3=2(2)+b
3=4+b
b=3+4
b=−1
​
Substitute ​m=2 b=−1 in the slope-intercept form, and we get
y=2x−1

Using a graphing calculator, the graph of the equation y=2x−1 is

The graph passes through the given points(2,3) and(0,−1). So, the equation is correct.

For the given graph
The slope is m=2
The y-intercept is b=−1

The equation of each line in the slope-intercept form will be y=2x−1

 

Given:

To find, find the slope and y-intercept

The slope of the line m\(=\frac{y_2-y_1}{x_2-x_1}\)

The slope-intercept form y=mx+b

Here​(x₁,y₁)=(0,2) , (x₂,y₂)=(4,−2)
​
By substituting, we get

​m\(=\frac{-2-2}{4-0}\)

\(=\frac{-4}{4}\)

=−1
​

In the slope-intercept form, we substitute m=−1 and take any point we get
​2=−1(0)+b
b=2
​
Substitute​ m=−1 in the slope-intercept form, and we get
b=2
​y=(−1)x+2
y=−x+2
​

Using a graphing calculator, the graph of the equation y=−x+2 is

The graph passes through the given points(0,2) and(4,−2). So, the equation is correct.

For the given graph
The Slope​ m=−1
The y-Intercept b=2

The equation of each line in the slope-intercept form will be y=−x+2

 

Given:

To find, find the slope and y-Intercept

The slope of the line m\(m=\frac{y_2-y_1}{x_2-x_1}\)

The slope-intercept form y=mx+b

Here​(x₁,y₁)=(−3,3)(x₂,y₂)=(3,−1)
​
By substituting, we get

​m\(=\frac{-1-3}{3-(-3)}\)

\(=\frac{-4}{6}\) \(=\frac{-2}{3}\)

In the slope-intercept form, we substitute m\(=\frac{-2}{3}\) and take any point we get

\(=\left(\frac{-2}{3}\right)(-3)+b\)
​
b=3−2

b=1
​
Substitute ​\(m=\frac{-2}{3}\) in the slope-intercept form, we get,

b=1

y\(=\frac{-2}{3} x\)+1

Using a graphing calculator, the graph of the equation y\(=\frac{-2}{3} x\)+1 is

The graph passes through the given points(−3,3) and(3,−1). So, the equation is correct.

For the given graph
The slope m\(=\frac{-2}{3}\)
The y-intercept b=1

The equation of each line in the slope-intercept form will be y\(=\frac{-2}{3} x+1\)

 

Given:

To find, find the slope and y-intercept
The slope of the line m \(=\frac{y_2-y_1}{x_2-x_1}\)
The slope-intercept form y=mx+b

Here​(x₁,y₁)=(4,0)(x₂,y₂)=(2,−1)
​
By substituting, we get

​m\(=\frac{-1-0}{2-4}\)

\(=\frac{-1}{-2}\) \(=\frac{1}{2}\)

In the slope-intercept form, we substitute m\(=\frac{1}{2}\) and take any point we get

​0=\(4\left(\frac{1}{2}\right)+b\)

b+2=0

b=−2
​
Substitute​ m\(=\frac{1}{2}\) in the slope-intercept form, we get

b=−2
​
y\(=\frac{1}{2} x\)+(-2)

y\(=\frac{x}{2}-2\)

For the given graph

The slope m\(=\frac{1}{2}\)

The y-Intercept b=−2

The equation of each line in the slope-intercept form will be y\(=\frac{x}{2}-2\)

Big Ideas Math Algebra 1 Chapter 4 Exercise 4.1 Solution

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 4 Writing Linear Functions Exercise 4.1 Page 97  Exercise 2  Answer

Given:

To find, the y-intercept of the line.
The y-intercept of the graph represents the point that crosses the y-axis.
In this graph, we can observe that the y-intercept is at y=20.

The y-intercept of the graph is y=20, this represents the initial cost of the smartphone plan.

 

Given:

To find, the slope of the line. Interpret the slope in the context.
The slope of the line m\(=\frac{y_2-y_1}{x_2-x_1}\)

Here we take two points from the graph
​(x₁,y₁)=(0,20)
(x₂,y₂)=(2000,80)
​
By substituting, we get

​m=\(\frac{80-20}{2000-0}\)

\(=\frac{60}{2000}\) \(=\frac{3}{100}\)

The slope of the line of the graph is m\(=\frac{3}{100}\), this represents the change in cost for a smartphone plan based on data usage.
​

Given:

To find, an equation that represents the cost as a function of data usage. The slope-intercept form y=mx+b

We get​ m\(=\frac{3}{100}\)
b=20
​By substituting, we get
y\(=\frac{3}{100}\)x+20
where x is the data usage.

The equation that represents the cost as a function of data usage will be y\(=\frac{3}{100}\)x+20

Writing Linear Functions Exercise 4.1 Big Ideas Math

Page 97  Exercise 3  Answer

​Given: The linear function

To find, the equation of the line
Here we solve it graphically.

For any graph of a linear function, we can determine the equation of the line by obtaining the slope and y-intercept.
Once those two values are identified, we can use the slope-intercept form of the function y=mx+b to write the equation of the line.

Let’s take the equation of the line y=−3x+7

 

We can write the equation of a line using the slope and the y-intercept For example the graph for the equation of the line y=−3x+7

 

Page 97   Exercise 4  Answer

Given:

To find, the equation of the line.

The slope of the line m\(=\frac{y_2-y_1}{x_2-x_1}\)

The slope-intercept form y=mx+b

We use two arbitrary points(x₁,y₁)=(6,9)(x₂,y₂)=(4,20)

​By substituting we get

​\(m=\frac{20-9}{4-6}\)

\(=\frac{-11}{2}\)

In the slope-intercept form, we substitute \(m=\frac{-11}{2}\)and take any point we get

​9\(=\left(\frac{-11}{2}\right)(6)+b\)

9=−33+b

b=33+9

b=42
​
Substitute​ m= \(\frac{-11}{2}\) in the slope-intercept form, we get

b=42
​
y=\(\frac{-11}{2}\)x+42

The graph for the equation y\(\frac{-11}{2} x+42\) will be

 

For the taken arbitrarily points,
The equation of the line y=\(\frac{-11}{2} x+42\)
The graph will be

Algebra 1 Student Journal Chapter 4 Exercise 4.1 Answers

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 4 Writing Linear Functions Exercise 4.1 Page 99  Exercise 1  Answer

Given: ​Slope=0

y−Intercept=9

To find, the equation of the line
The slope-intercept form of the line is y=mx+b

Here Slope m=0
And y-Intercept b=9
By substituting, we get
​y=0(x)+9
y=9
​
The equation of the line for the given slope and y-intercept will be y=9.

 

Page 99  Exercise 2  Answer

Given: ​Slope=−1

y−Intercept=0
​
To find, the equation of the line
The slope-intercept form of the line y=mx+b

Here slope m=−1

The y-Intercept b=0

By substituting, we get

​y=(−1)x+b

y=−x
​

The equation of the line for the given slope and y-intercept will be y=−x

 

Page 99 Exercise 3  Answer

Given:​ Slope=2

y−Intercept=−3
​
To find, the equation of the line
The slope of the line y=mx+b

Here slope m=2

The y-intercept b=−3

By substituting, we get

​y=2x+(−3)

y=2x−3
​

The equation of the line for the given slope and y-intercept will be y=2x−3

 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 4 Writing Linear Functions Exercise 4.1 Page 99  Exercise 4  Answer

Given:​ Slope=−3

y−Intercept=7

To find, the equation of the line
The slope-intercept form of the line is y=mx+b

Here Slope m=−3

The y-Intercept b=7

By substituting, we get

y=−3x+7

The equation of the line for the given slope and y-intercept will be y=−3x+7

 

Page 99  Exercise 5  Answer

Given: ​Slope=4

y−Intercept=−2

To find, the equation of the line
The slope-intercept form of the line is y=mx+b

Here slope m=4
The y-intercept b=−2
By substituting, we get
​y=4x+(−2)
y=4x−2
​

The equation of the line for the given slope and y-intercept will be y=4x−2

Big Ideas Math Writing Linear Functions Exercise 4.1 Help

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 4 Writing Linear Functions Exercise 4.1 Page 99  Exercise 6  Answer

Given:​ Slope\(=\frac{1}{3}\)

y−Intercept t=2

To find, the equation of the line
The slope of the line y=mx+b

Here slope m\(=\frac{1}{3}\)

The y-intercept b=2

By substituting, we get

​y=(\(\frac{1}{3}\))x+2

y\(=\frac{x}{3}+2\)
​

The equation of the line for the given slope and y-intercept will be y \(=\frac{x}{3}+2\)

 

Page 99  Exercise 7  Answer

Given: graph with two points (−1,3),(0,−1) on the line is

To write an equation of a line in slope-intercept form.

We’ll find the slope of the line using the formula of finding the slope of a line passing through two points m=\(\frac{y_2-y_1}{x_2-x_1}\)

Then we’ll write the equation in point-slope form using the formula y−y₁ =m(x−x₁) and simplify it to get a slope-intercept form.

The slope of line passing through the points (−1,3),(0,−1) is

​m\(=\frac{-1-3}{0-(-1)}\)

\(=\frac{-4}{1}\)

=−4
​

We have slope m=−4 and a point on the line (−1,3)

The equation of the line in point-slope form can be written as
​y−3=(−4)(x−(−1))
⇒ y−3=(−4)(x+1)
⇒ y−3=−4x−4
⇒ y=−4x−1
​
So, the slope-intercept form of a line is y=−4x−1 with
​m=−4
c=−1

fn-a
The equation of a line in slope-intercept form is y=−4x−1 with
​m=−4
c=−1

 

Page 99  Exercise 8  Answer

Given: graph with two points (2,4),(0,0) on the line

To write an equation of a line in slope-intercept form.

We’ll find the slope of the line using the formula of finding the slope of a line passing through two points m\(m=\frac{y_2-y_1}{x_2-x_1}\)

Then we’ll write the equation in point-slope form using the formula y−y₁
=m(x−x₁)and simplify it to get a slope-intercept form.

The slope of a line passing through the points (2,4),(0,0) can be found using the formula as-
​
m\(=\frac{0-4}{0-2}\)
​
\(=\frac{-4}{-2}\)

=2

We have m=2 and a point (2,4)
The equation of the line in point-slope form can be written as
​y−4=2(x−2)
⇒y−4=2x−4
⇒y=2x
​
So, the slope-intercept form of a line is y=2x n with
​m=2
c=0
​

The equation of a line in slope-intercept form is y=2x with
​m=2
c=0

Chapter 4 Exercise 4.1 Step-By-Step Solutions Big Ideas Math

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 4 Writing Linear Functions Exercise 4.1 Page 99  Exercise 9  Answer

Given: graph with two points (2,3),(0,1) on the line

To write an equation of line in slope-intercept form.

We’ll find the slope of the line using the formula of finding slope of a line passing through two points m\(=\frac{y_2-y_1}{x_2-x_1}\)

Then we’ll write the equation in point-slope form using the formula y−y₁=m(x−x₁) and simplify it to get a slope-intercept form.

The slope of line passing through the points (2,3),(0,1) can be found using the formula as

​m\(=\frac{1-3}{0-2}\)

\(=\frac{-2}{-2}\)

=1

We have, m=1 and a point (2,3)
The equation of the line in point-slope form can be written as
​y−3=1(x−2)
⇒y−3=x−2
⇒y=x+1
​
So, the slope-intercept form of line is y=x+1 with
​m=1
c=1

The equation of line in slope-intercept form is y=x+1 with
​m=1
c=1
​

 

​Page 99  Exercise 10  Answer

Given: graph with two points (0,5),(3,−4) on the line

To write an equation of a line in slope-intercept form.

We’ll find the slope of the line using the formula of finding the slope of a line passing through two points m\(=\frac{y_2-y_1}{x_2-x_1}\)

Then we’ll write the equation in point-slope form using the formula y−y₁
=m(x−x₁) and simplify it to get a slope-intercept form.

The slope of line passing through the points (0,5),(3,−4) can be found using the formula as

​m\(=\frac{-4-5}{3-0}\)

\(=\frac{-9}{3}\)

=−3
​

We have m=−3 and a point (0,5)
The equation of the line in point-slope form can be written as
​y−5=(−3)(x−0)
⇒y−5=−3x
⇒y=−3x+5

So, the slope-intercept form of a line is y=−3x+5 with
​m=−3
c=5
​

The equation of a line in slope-intercept form is y=−3x+5 with
​m=−3
c=5

​Exercise 4.1 Big Ideas Math Algebra 1 Guide 

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 4 Writing Linear Functions Exercise 4.1Page 99  Exercise 11  Answer

Given: graph with two points (−2,−3),(0,−2) on the line is

To write an equation of a line in slope-intercept form.

We’ll find the slope of the line using the formula of finding the slope of a line passing through two points m\(m=\frac{y_2-y_1}{x_2-x_1}\)

Then we’ll write the equation in point-slope form using the formula y−y₁=m(x−x₁)
and simplify it to get a slope-intercept form.

The slope of a line passing through the points (−2,−3),(0,−2) can be found using the formula as

m\(=\frac{-2-(-3)}{0-(-2)}\)

\(=\frac{-2+3}{2}\) \(=\frac{1}{2}\)

We have, m\(=\frac{1}{2}\) and a point (−2,−3)

The equation of the line in point-slope form can be written as

​y−(−3)=\(\frac{1}{2}\)(x−(−2))

⇒y+3=\(\frac{1}{2}\)(x+2)

⇒y=\(\frac{1}{2}\)x+\(\frac{1}{2}\)(2)-3

⇒y=\(\frac{1}{2}\)x+1-3

⇒y=\(\frac{1}{2}\)-2

So, the slope-intercept form of a line is y=\(\frac{1}{2}\)-2 with ​m\(=\frac{1}{2}\)

c=−2
​

The equation of a line in slope-intercept form is y=\(\frac{1}{2}\)-2
with\(=\frac{1}{2}\)
c=−2
​

Page 99  Exercise 12  Answer

Given: graph with two points (0,3),(4,0) on the line

To write an equation of a line in slope-intercept form.

We’ll find the slope of the line using the formula of finding the slope of a line passing through two points m \(=\frac{y_2-y_1}{x_2-x_1}\).

Then we’ll write the equation in point-slope form using the formula y−y₁ =m(x−x₁) and simplify it to get a slope-intercept form.

The slope of a line passing through the points (0,3),(4,0) can be found using the formula as

​m\(=\frac{0-3}{4-0}\)

\(=-\frac{3}{4}\)

We have, m\(=\frac{-3}{4}\) and a point (0,3)
The equation of the line in point-slope form can be written as

​y−3\(=\left(\frac{-3}{4}\right)(x-0)\)

⇒ y−3=\(\frac{-3}{4} x\)

⇒ y\(=\frac{-3}{4} x+3\)

So, the slope-intercept form of a line is y=\(\frac{-3}{4} x+3\) with

​m\(=\frac{-3}{4}\)

c=3
​
The equation of a line in slope-intercept form is y=\(\frac{-3}{4} x+3\) with

​\(=\frac{-3}{4}[latex]

c=3

 

Page 100  Exercise 13  Answer

Given: Two points on the line (3,−1),(8,4)

To find equation of the line passing through the given two points.

Using the formula of equation of line passing through two points, we can find the required equation.

By using the formula of the equation of the line passing through two points,  y−y₁=[latex]\frac{y_2-y_1}{x_2-x_1}\)(x−x₁)

We can calculate the equation of a line passing through the points (3,−1),(8,4) as

​y−(−1)\(=\frac{4-(-1)}{8-3}\)(x−3)

⇒ y+1=\(\frac{5}{5}\) (x−3)

⇒ y+1=x−3

⇒ y−x+4=0
​
The equation of a line passing through the points (3,−1),(8,4) is y−x+4=0.

How To Solve Exercise 4.1 Big Ideas Math

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 4 Writing Linear Functions Exercise 4.1 Page 100  Exercise 14  Answer

Given: two points on the line (2,1),(3,5)

To find equation of the line passing through the given two points

Using the formula of equation of line passing through two points, we can find the required equation.

By using the formula of the equation of the line passing through two points- y−y₁\(=\frac{y_2-y_1}{x_2-x_1}\)(x−x₁) we can calculate the equation of the line passing through the points (2,1),(3,5) as

​y−1\(=\frac{5-1}{3-2}\)(x−2)

⇒ y−1=4(x−2)

⇒ y−1=4x−8

⇒ y−4x+7=0
​

The equation of line passing through the points (2,1), and (3,5) is y−4x+7=0

 

Page 100  Exercise 16  Answer

Given: two points on the line (−3,−2),(−4,−1)

To find equation of the line passing through the given two points.

Using the formula of equation of line passing through two points, we can find the required equation.

By using the formula of equation of the line passing through two points- y−y₁
\(=\frac{y_2-y_1}{x_2-x_1}\)(x−x₁) we can calculate the equation of line passing through the points (−3,−2),(−4,−1) as

​y−(−2)\(=\frac{-1-(-2)}{-4-(-3)}\)(x−(−3))

⇒ y+2\(=\frac{-1+2}{-4+3}\)(x+3)

⇒ y+2=\(\frac{1}{-1}\)(x+3)

⇒ y+2=−(x+3)

⇒ y+2=−x−3

⇒ y+x+5=0

The equation of the line passing through the points (−3,−2),(−4,−1) is y+x+5=0.
​

 

Page 100  Exercise 17  Answer

Given: Two coordinates are given​(8,0),(0,8)
​
Find, Equation of the line that passes through given coordinates. To find the equation of the line, first, find out the slope of the line using the formula ​m\(=\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\).

Then to find out the equation of the line using the slope-intercept form we will use the formula ​y=mx+c.

For Example, two points given are ​(7,5),(−9,5)
​
To find out the slope of the line we will substitute the values of the x,y coordinates in the formula ​mm\(=\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\)

So m\(=\frac{5-5}{-9-7}\)

​\(=\frac{0}{-16}\)

=0
​

To find out the equation of the line we use the formula y=mx+c, substitute the value of mand one of the coordinates in the above formula, and get the value of c.

Then get the equation by substituting the value of m,c in the formula.

To calculate the slope we’ll use the formula, ​m\(=\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\)

​m\(=\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\)

​​\(=\frac{(8-0)}{(0-8)}\)

\(=\frac{8}{-8}\)

=−1
​
​Using the formula of slope-intercept, ​y=mx+c, along with the values of m and point (0,8)we calculate c as follows
​y=mx+c
8=(−1)×(0)+c
8=0+c
c=8
​
Substitute the value of m,c in the equation y=mx+c
​y=mx+c
y=(−1)×x+8
y=−x+8
​

The equation of the line that passes through the given points,(8,0),(0,8) is y=−x+8

Big Ideas Math Chapter 4 Exercise 4.1 Walkthrough

 Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 4 Writing Linear Functions Exercise 4.1Page 100  Exercise 18  Answer

Two coordinates are provided(−1,7),(2,−5).

Find, the line’s equation that passes through the given coordinates.

To calculate the line’s equation, first determine the line’s slope using the formula.​m\(=\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\)

​
Then, using the slope-intercept form, we’ll use the formula to obtain the line’s equation y=mx+c.

For instance, consider the following two points:(4,7),(6,13) We’ll use the formula m=\(\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\)to get the slope of the line by substituting the coordinate values x,y.

So,​m \(=\frac{13-7}{6-4}\)

​\(=\frac{6}{2}\)

=3
​

We apply the formula y=mx+c to determine the equation of the line,

Get the value of c by substituting the value of m and one of the coordinates in the given formula.

Substituting the value of m,c into the formula yields the equation.

To calculate the slope we’ll use the formula

​m=\(\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\)

​​m=\(\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\)

\(=\frac{(-5-7)}{(2-(-1))}\) \(=\frac{-12}{3}\)

=−4
​

Using the formula of slope-intercept, ​y=mx+c, along with the values of m and point(−1,7) we calculate as follows
​y=mx+c
7=(−4)×(−1)+c
7=4+c
c=3
​

Substitute the value of m,c in the equationy=mx+c
​y=mx+c
y=(−4)×x+3
y=−4x+3
​

Equation of the line that passes through the given points(−1,7),(2,−5) , is y=−4x+3

 

Page 100  Exercise 20  Answer

Given: f(−5)=5,f(5)=15

Find a linear function with the values you’ve provided.

Using the function provided, find the coordinates.

To calculate the line’s equation, first determine the line’s slope using the formula.m\(=\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\)

Then, using the slope-intercept form, we’ll use the formula y=mx+c to obtain the line’s equation.

To find the coordinates, follow these steps:

The y-values are the value of f(x), thus we have the points.(−5,5),(5,15)

To calculate the formula we’ll use the formula,m=\(\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\)

m=\(\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\)
​
\(=\frac{(15-5)}{(5-(-5))}\)

\(=\frac{10}{10}\)

=1
​

Using the formula of slope-intercept,y=mx+c , along with the values of m and point(5,15) , we calculate c as follows,
​y=mx+c
15=(1)×(5)+c
15=5+c
c=10
​

Substitute the value of m,c in the equation y=mx+c,
y=mx+c
y=(1)×x+10
y=x+10
​

Equation of the line that passes through the given points,(−5,5),(5,15) is y=x+10

 

Page 100  Exercise 22  Answer

Given: f(2)=6, f(7)=−4

Find, Linear function with the given values.

Find the coordinates using the Function given.

To find the equation of the line, first, find out the slope of the line using the formula m=\(\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\).

Then to find out the equation of the line using the slope-intercept form we will use the formula y=mx+c.

To find the coordinates:

The value of  f(x)
would be the y-values, hence we have the points(2,6),(7,−4)

To calculate the slope we’ll use the formula,m=\(\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\)

m=\(\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\)

\(=\frac{(-4-6)}{(7-2)}\)

​\(=\frac{-10}{5}\)

=-2

Using the formula of slope-intercept, y=mx+c, along with the values of m and point(7,−4), we calculate c as follows,
​y=mx+c
−4=−2×7+c
−4=−14+c
c=10
​

Substitute the value of m,c in the equationy=mx+c,
​y=mx+c
y=−2×x+10
y=2x+10
​

Equation of the line that passes through the given points,(2,6),(7,−4) is y=2x+10

Algebra 1 Exercise 4.1 Explanation Big Ideas Math

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 4 Writing Linear Functions Exercise 4.1 Page 100  Exercise 23  Answer

Given: f(−2)=−2, f(4)=10

Find a linear function with the values you’ve provided.

Using the function provided, find the coordinates.

To calculate the line’s equation, first determine the line’s slope using the formula.m\(=\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\)

Then, using the slope-intercept form, we’ll use the formula y=mx+c to obtain the line’s equation.

To find the coordinates:

The y-values are the value of f(x), thus we have the points.(−2,−2),(4,10)

To calculate the formula we’ll use the formula,m\(=\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\)

m\(=\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\)

\(=\frac{(10-(-2))}{(4-(-2))}\) \(=\frac{12}{6}\)

=2
​

Using the formula of slope-intercept,y=mx+c, along with the values of m and point(4,10), we calculate c as follows
​y=mx+c
10=2×4+c
10=8+c
c=2
​

Substitute the value of m,c in the equation y=mx+c
​y=mx+c
y=2×x+2
y=2x+2
​

Equation of the line that passes through the given points, (−2,−2),(4,10)​ is y=2x+2.

 

Page 100  Exercise 24  Answer

Given: f(4)=0,f(2)=8

Find, Linear function with given values.

Find the coordinates using the function given.

To find the equation of the line, first, find out the slope of the line using the formula m\(=\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\).

Then to find out the equation of the line using the slope-intercept form we will use the formula y=mx+c.

To find the coordinates:

The value of f(x)
would be the y-values, hence we have the points (4,0),(2,8)

To calculate the slope we’ll use the formula,m \(=\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\)

 

m\(=\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\)

 

\(=\frac{(8-0)}{(2-4)}\)

 

\(=\frac{8}{-2}\)

 

=−4
​

Using the formula of slope-intercept,y=mx+c, along with the values of m and point(2,8), we calculate c as follows
​y=mx+c
8=−4×2+c
8=−8+c
c=16
​

Substitute the value of m,c in the equation y=mx+c,
​y=mx+c
y=−4×x+16
y=−4x+16
​

The equation of the line that passes through the given points, (4,0),(2,8) is y=−4x+16

Big Ideas Math Student Journal Exercise 4.1 Examples