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Mathematics 12 Online
OpenStudy (anonymous):

f(x)=1/2(x=3) find domain and sketch the graph

OpenStudy (calculusfunctions):

\[f(x)=\frac{ 1 }{ 2(x -3) }\]Is this the correct equation or is it\[f(x)=\frac{ 1 }{ 2 }(x -3)\]Before I help, I need to know which one it is because it's not very clear from your presentation of the equation.

OpenStudy (anonymous):

2nd one

OpenStudy (calculusfunctions):

OK, so then what type of of function is\[f(x)=\frac{ 1 }{ 2 }(x -3)\] a). linear b). quadratic c). reciprocal d). cubic

OpenStudy (anonymous):

linear ? ? i guess

OpenStudy (calculusfunctions):

I'm going to teach you, not just give you the answers, so could you please answer my question? If you don't know the answer and you need a hint or help, then just say so.

OpenStudy (calculusfunctions):

Yes, it is linear. Good!

OpenStudy (anonymous):

thanks man

OpenStudy (calculusfunctions):

Now can you tell me what you know about linear functions?

OpenStudy (anonymous):

cant recall. :(

OpenStudy (calculusfunctions):

Can you tell me the name of the graph of a linear function?

OpenStudy (anonymous):

i dont know

OpenStudy (calculusfunctions):

Look at the word linear. What other word do you see within the word "LINEAR".

OpenStudy (anonymous):

near?

OpenStudy (anonymous):

straight line ???

OpenStudy (calculusfunctions):

Yes, LINE! Good!

OpenStudy (calculusfunctions):

Have you ever seen or heard of the slope-intercept form of the equation of a line?

OpenStudy (anonymous):

nope

OpenStudy (anonymous):

|dw:1350031122437:dw|

OpenStudy (anonymous):

is this the answer?

OpenStudy (calculusfunctions):

A linear function is a first degree polynomial function. First degree because the highest power (exponent) of x is one. The graph of a linear function is a straight line. The equation of a diagonal (oblique) line in slope-intercept form is given by\[y =mx +b\]where m is the slope (ratio of rise to run) of the line and b is the y-intercept of the line.

OpenStudy (calculusfunctions):

Your diagram is flawed.

OpenStudy (anonymous):

we dont use slope

OpenStudy (calculusfunctions):

Do you use the word gradient?

OpenStudy (anonymous):

this answer is suppose to be really simple and straight forward..

OpenStudy (anonymous):

yes

OpenStudy (calculusfunctions):

Yes but do you just want me to answer your question, or would you like to learn something here that will also help you later?

OpenStudy (anonymous):

thats good that you teach me. well problem is i dont know wether it is correct. Our teacher said by using function f(0) we findo out two points and then by solving the equation of 1/2((x-3)

OpenStudy (anonymous):

I did how he said but using slope will be easy

OpenStudy (calculusfunctions):

OK fine, do you know the significance of f(0)?

OpenStudy (calculusfunctions):

Do you know what f(x) means?

OpenStudy (anonymous):

then subsitute x to the functiomn so it will be 1/2(0+3)

OpenStudy (anonymous):

so it is 3/2 as ans

OpenStudy (anonymous):

tht means x=) and y=1.5

OpenStudy (anonymous):

srry x=0

OpenStudy (calculusfunctions):

Yes, f(0) means you substitute x = 0 into the equation, but what does that mean when you find f(0)?

OpenStudy (calculusfunctions):

Yes it is 3/2 = 1.5 but what does that mean?

OpenStudy (anonymous):

that y=1.5

OpenStudy (calculusfunctions):

f(0) = 3/2 = 1.5 but what does it mean?

OpenStudy (anonymous):

when x is 0 the y point lies on 1.5

OpenStudy (calculusfunctions):

Does it mean that you found a). the slope b) the y-intercept of the line c). the x-intercept of the line d). an arbitrary point Please provide the best response.

OpenStudy (anonymous):

y intercept

OpenStudy (calculusfunctions):

Correct! Good!

OpenStudy (calculusfunctions):

Now how do you find the x-intercept?

OpenStudy (anonymous):

yes x=-3

OpenStudy (anonymous):

by solving x+3/2=0

OpenStudy (calculusfunctions):

Does it? Are you sure? Is that your final answer?

OpenStudy (anonymous):

so it is x intercept and we founf y intercept with two points and then we will draw a straight line

OpenStudy (calculusfunctions):

But you still didn't tell me what your final response for the x-intercept is.

OpenStudy (anonymous):

thank you

OpenStudy (calculusfunctions):

No we're not done yet.

OpenStudy (anonymous):

ahh i have to go now

OpenStudy (anonymous):

how about later but still i dnt know what i missed

OpenStudy (calculusfunctions):

You still haven't found the domain yet.

OpenStudy (anonymous):

ahh i think all number is real

OpenStudy (calculusfunctions):

And your graph was wrong.

OpenStudy (calculusfunctions):

Yes, The domain is all real numbers for x. But your graph is still wrong. Do you know why?

OpenStudy (anonymous):

im trying

OpenStudy (anonymous):

|dw:1350032824915:dw|

OpenStudy (anonymous):

correct right

OpenStudy (calculusfunctions):

Ask yourself if you plotted the x-int: (3, 0) and the y-int: (0, -1.5) correctly. By the way, the x-int is at (3, 0) and NOT (-3, 0). Also I was wrong to say that your y-int was correct because it is NOT. The y-int is at (0, -1.5) NOT (0, 1.5). Do you understand?

OpenStudy (calculusfunctions):

Plot the graph again.

OpenStudy (anonymous):

ahh ok so if it is x interxept y=0 and y intercept x=0

OpenStudy (calculusfunctions):

Eureka!

OpenStudy (calculusfunctions):

Excellent! So now could you please plot the graph again?

OpenStudy (anonymous):

|dw:1350033079174:dw|

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