# 4.1 Linear functions  (Page 10/27)

 Page 10 / 27
$y=2x+4$

Given a graph of linear function, find the equation to describe the function.

1. Identify the y- intercept of an equation.
2. Choose two points to determine the slope.
3. Substitute the y- intercept and slope into the slope-intercept form of a line.

## Matching linear functions to their graphs

Match each equation of the linear functions with one of the lines in [link] .

Analyze the information for each function.

1. This function has a slope of 2 and a y -intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. We can use two points to find the slope, or we can compare it with the other functions listed. Function $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ has the same slope, but a different y- intercept. Lines I and III have the same slant because they have the same slope. Line III does not pass through $\text{\hspace{0.17em}}\left(0,\text{3}\right)\text{\hspace{0.17em}}$ so $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ must be represented by line I.
2. This function also has a slope of 2, but a y -intercept of $\text{\hspace{0.17em}}-3.\text{\hspace{0.17em}}$ It must pass through the point $\text{\hspace{0.17em}}\left(0,-3\right)\text{\hspace{0.17em}}$ and slant upward from left to right. It must be represented by line III.
3. This function has a slope of –2 and a y- intercept of 3. This is the only function listed with a negative slope, so it must be represented by line IV because it slants downward from left to right.
4. This function has a slope of $\text{\hspace{0.17em}}\frac{1}{2}\text{\hspace{0.17em}}$ and a y- intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. Lines I and II pass through $\text{\hspace{0.17em}}\left(0,\text{3}\right),$ but the slope of $\text{\hspace{0.17em}}j\text{\hspace{0.17em}}$ is less than the slope of $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ so the line for $\text{\hspace{0.17em}}j\text{\hspace{0.17em}}$ must be flatter. This function is represented by Line II.

Now we can re-label the lines as in [link] .

## Finding the x -intercept of a line

So far we have been finding the y- intercepts of a function: the point at which the graph of the function crosses the y -axis. Recall that a function may also have an x -intercept , which is the x -coordinate of the point where the graph of the function crosses the x -axis. In other words, it is the input value when the output value is zero.

To find the x -intercept, set a function $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ equal to zero and solve for the value of $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ For example, consider the function shown.

$f\left(x\right)=3x-6$

Set the function equal to 0 and solve for $\text{\hspace{0.17em}}x.$

$\begin{array}{ccc}\hfill 0& =& 3x-6\hfill \\ \hfill 6& =& 3x\hfill \\ \hfill 2& =& x\hfill \\ \hfill x& =& 2\hfill \end{array}$

The graph of the function crosses the x -axis at the point $\text{\hspace{0.17em}}\left(2,\text{0}\right).$

Do all linear functions have x -intercepts?

No. However, linear functions of the form $\text{\hspace{0.17em}}y=c,$ where $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ is a nonzero real number are the only examples of linear functions with no x-intercept. For example, $\text{\hspace{0.17em}}y=5\text{\hspace{0.17em}}$ is a horizontal line 5 units above the x-axis. This function has no x-intercepts, as shown in [link] .

## x -intercept

The x -intercept of the function is value of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}f\left(x\right)=0.\text{\hspace{0.17em}}$ It can be solved by the equation $\text{\hspace{0.17em}}0=mx+b.$

## Finding an x -intercept

Find the x -intercept of $\text{\hspace{0.17em}}f\left(x\right)=\frac{1}{2}x-3.$

Set the function equal to zero to solve for $\text{\hspace{0.17em}}x.$

$\begin{array}{ccc}\hfill 0& =& \frac{1}{2}x-3\hfill \\ \hfill 3& =& \frac{1}{2}x\hfill \\ \hfill 6& =& x\hfill \\ \hfill x& =& 6\hfill \end{array}$

The graph crosses the x -axis at the point $\text{\hspace{0.17em}}\left(6,\text{0}\right).$

Find the x -intercept of $\text{\hspace{0.17em}}f\left(x\right)=\frac{1}{4}x-4.$

## Describing horizontal and vertical lines

There are two special cases of lines on a graph—horizontal and vertical lines. A horizontal line    indicates a constant output, or y -value. In [link] , we see that the output has a value of 2 for every input value. The change in outputs between any two points, therefore, is 0. In the slope formula, the numerator is 0, so the slope is 0. If we use $\text{\hspace{0.17em}}m=0\text{\hspace{0.17em}}$ in the equation $\text{\hspace{0.17em}}f\left(x\right)=mx+b,$ the equation simplifies to $\text{\hspace{0.17em}}f\left(x\right)=b.\text{\hspace{0.17em}}$ In other words, the value of the function is a constant. This graph represents the function $\text{\hspace{0.17em}}f\left(x\right)=2.$

Cos45/sec30+cosec30=
Cos 45 = 1/ √ 2 sec 30 = 2/√3 cosec 30 = 2. =1/√2 / 2/√3+2 =1/√2/2+2√3/√3 =1/√2*√3/2+2√3 =√3/√2(2+2√3) =√3/2√2+2√6 --------- (1) =√3 (2√6-2√2)/((2√6)+2√2))(2√6-2√2) =2√3(√6-√2)/(2√6)²-(2√2)² =2√3(√6-√2)/24-8 =2√3(√6-√2)/16 =√18-√16/8 =3√2-√6/8 ----------(2)
exercise 1.2 solution b....isnt it lacking
I dnt get dis work well
what is one-to-one function
what is the procedure in solving quadratic equetion at least 6?
Almighty formula or by factorization...or by graphical analysis
Damian
I need to learn this trigonometry from A level.. can anyone help here?
yes am hia
Miiro
tanh2x =2tanhx/1+tanh^2x
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)=cotb ... pls some one should help me with this..thanks in anticipation
f(x)=x/x+2 given g(x)=1+2x/1-x show that gf(x)=1+2x/3
proof
AUSTINE
sebd me some questions about anything ill solve for yall
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)= cotb
favour
how to solve x²=2x+8 factorization?
x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN)
Manifoldee
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
I mean x²=2x+8 by factorization method
Kristof
I think x=-2 or x=4
Kristof
x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia
Mohamed
i am in
Cliff
hii
Amit
how are you
Dorbor
well
Biswajit
can u tell me concepts
Gaurav
Find the possible value of 8.5 using moivre's theorem
which of these functions is not uniformly cintinuous on (0, 1)? sinx
helo
Akash
hlo
Akash
Hello
Hudheifa
which of these functions is not uniformly continuous on 0,1