Let f be a linear function where f(2)= -5 and f(-3)=1. Find f(x).
HOW TO DO?!?!?!?!?!

- anonymous

Let f be a linear function where f(2)= -5 and f(-3)=1. Find f(x).
HOW TO DO?!?!?!?!?!

- Stacey Warren - Expert brainly.com

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- schrodinger

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- RadEn

the general of a linear function is
f(x) = ax + b (with a and b are the constant)
now given f(2) = -5, it means
a(-2) + b = -5 or
-2a + b = -5 ........ (1)
then known the value of f(-3) = 1, it means
a(-3) + b = 1 or
-3a + b = 1 ........ (2)
now you have to solve for a and b respectively. (in the 1st and the 2nd eq)
you can solve them by elimination and subtitution method

- RadEn

got it ?

- anonymous

I still do not understand sir.

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## More answers

- anonymous

They gave u two points that are on the line, so you can calculate the slope

- anonymous

do u know how?

- anonymous

what are the points..?

- anonymous

They told u that f(2)= -5, which means when you plug in 2 as x, you get -5 as y, so the point (2,-5) is on the line

- anonymous

okay and the other point is (-3,1), right?

- anonymous

yep

- anonymous

u know the formula for slope?

- anonymous

You mean the point slope formula?

- anonymous

You're gonna need the slope to use that formula

- anonymous

have u ever heard of rise over run?

- anonymous

ohh i know is it (y-y1)/( x-x1)?

- anonymous

Ya thats right, so find the slope and then you can use the slope and point on the line to find the point slope form

- RadEn

the slope formula doesnt works here. like i said above we have 2 equations :
-2a + b = -5
-3a + b = 1
now, can you solve for a and b, first ?

- anonymous

Wait, i thought i was suppose to find the slope and then point slope formula to find the point slope form. What does a and b have to do with this?

- RadEn

they are the constant, and again like i said above :
the general of a linear function is f(x) = ax + b
so, we need the values of a and b, after you geot them then just plug into equation above
|dw:1377653399115:dw|

- RadEn

got *

- anonymous

Ohh I see, so I use elimination method?

- RadEn

yeah, hold on :)

- anonymous

a=1 b=0

- anonymous

I cancelled out b.

- RadEn

subtract direct both equations, you will get :
-2a + b = -5
-3a + b = 1
-------------- (-)
a = -6

- anonymous

ohhh. so what to do after that

- RadEn

subtitution the value of a = -6 into one of equation above, let we take the 1st
-2a + b = 1
-2(-6) + b = 1
-12 + b = 1
add by 12 both sides, we get
-12 + b + 12 = 1 + 12
b = 13

- anonymous

alrighty, but the question ask to find f(x).

- RadEn

now, plug the values of a and b to f(x)
|dw:1377654643235:dw|

- anonymous

oh so when it says find f(x), it's asking for an equation?

- RadEn

yes, actually you have to find the values of the constants unknown then solve them and finally subtitute again them into that equation

- anonymous

what...

- anonymous

the answer is f(x)=-6x+13, right!?!?!?

- RadEn

wait. looks i make a mistake above

- anonymous

What mistake did you make?

- RadEn

-2a + b = 1
-2(-6) + b = 1
12 + b = 1
subtract by 12 both sides, we get
12 + b - 12 = 1 - 12
b = -11
sorry, that's my correction

- anonymous

alright so, FINAL ANSWER is f(x)=-6x-11. RIGHT?!?!

- RadEn

my mistake in -2 * -6 = -12, that should be 12.
yes, it is f(x) = -6x - 11

- anonymous

alrighty thanks man.

- RadEn

you're welcome :)

- RadEn

wait. the totally i mstake in calculation again above, so sorry

- RadEn

the first equation should
f(2) = -5,
a(2) + b = -5 or
2a + b = -5 ........ (1)

- RadEn

the 2nd equ already be correct
f(-3) = 1,
a(-3) + b = 1 or
-3a + b = 1 ........ (2)

- anonymous

what does ......(1) mean?

- theEric

Please ignore this, @izhangy , as it is not part of the solution! You are on the right track!
@RadEn and @PlsHaveMercyLilB , both methods are correct! The best option would probably whatever is easiest for izhangy or whatever the teacher is teaching. RadEn, you are looking to build and solve a system of equations. But do not discredit PlsHaveMercyLilB's solution, as \(\dfrac{y_2-y_1}{x_2-x_1}=m\)\(=\dfrac{f(2)-f(-3)}{2-(-3)}\)\(=\dfrac{(-5)-1}{2-(-3)}\)\(=\dfrac{-6}{1}=-6=m\)
Then you can make one point to be a very general \((x,y)\). So \(\dfrac{y_2-y_1}{x_2-x_1}=m\)\(=\dfrac{f(x)-f(-3)}{x-(-3)}\)
which \(=\dfrac{y-f(-3)}{x-(-3)}=m\) when I say \(y=f(x)\).
So, now we use algebra to solve for \(y\).
\(y-f(-3)\\=m(x-(-3))\\=m(x+3)\\=mx+3m\)
Then, after adding \(f(-3)=1\) to both sides, \(y=mx+3m+f(-3)=mx+3m+1\).
Since we said \(m=6\), \(\ \ y=6x+3(6)+1\\\quad=6x+18+1\\\quad =3x+19\)

- theEric

Does my work look good to you all?
Good work, everyone! :)

- anonymous

wait so you got f(x)= 3x+19?

- theEric

I did, but you'll have to check my work to see if it was right, or if what I got is actually what we're trying to get.
:)

- RadEn

that mean as the first equation,
2a + b = -5
-3a + b = 1
------------ (-)
5a = -6
a = -6/5
subtitute a= -6/5 to the first equation :
2a + b = -5
2 * (-6/5) + b = -5
-12/5 + b = -5
b = 12/5 - 5
b = (12 - 25)/5 = -13/5
so, f(x) = ax + b = (-6/5)x - 13/5

- RadEn

nah, we got the different of the result @theEric . hahaha..

- theEric

I see that! Haha!

- RadEn

just check, is it be right that f(-2) = -5 and f(-3) = 1

- anonymous

ok so who is right?

- theEric

@izhangy I'm checking with the other method, but it's up to YOU to find the answer. We can only be resources! :)

- RadEn

f(x) = ax + b = (-6/5)x - 13/5
let's check, would be satisfied by using trial error of the values f(-2) = -5 and f(-3) = 1.
first, PROVE f(-2) = -5
f(x) = (-6/5)x - 13/5
So, f(2) = (-6/5)(2) - 13/5 = -12/25 - 13/5 = -25/5 = -5 (IT IS RIGHT)
then
PROVE f(-3)=1
f(x) = (-6/5)x - 13/5
So, f(-3) = (-6/5)(-3) - 13/5 = 18/25 - 13/5 = 5/5 = 1 (IT IS RIGHT TOO)

- anonymous

K MAN, i believe you.

- anonymous

thanks all.

- RadEn

@theEric , do you agree with me, hehe :)

- anonymous

2a + b = -5
2 * (-6/5) + b = -5
-12/5 + b = -5
b = 12/5 - 5
b = (12 - 25)/5 = -13/5
okat wait a minute. @Raden how did u get from b= 12/5 - 5 to b = (12 - 25)/5 = -13/5.
Where did u get the 25 from?

- RadEn

|dw:1377657489398:dw|

- anonymous

ok

- theEric

\(f(2)= -5\) and \(f(-3)=1\), where \(f(x)\) is a linear function in the form of \(f(x)=mx+b\). Now, \(m\) and \(b\) are constant, and we can find them.
\(-5=m(2)+b\)
\(1=m(-3)+b\)
What is \(m\)?
\(-5=m(2)+b\implies m(2)=-5-b\implies m=\dfrac{-5-b}{2}\). Well, I expected it to be in terms of \(b\). But we can substitute \(m\) into the other equation.
\(1=m(-3)+b\\\implies 1=\dfrac{-5-b}{2}(-3)+b\\\qquad=\dfrac{15+3b}{2}+b\\\qquad=\dfrac{15}{2}+\dfrac{3b}{2}+b\\\qquad =\dfrac{15}{2}+b\left(\dfrac{3}{2}+1\right)=1\)
\(\implies b\left(\dfrac{3}{2}+1\right)=1-\dfrac{15}{2}\\\implies b=\dfrac{1-\dfrac{15}{2}}{\dfrac{3}{2}+1}=\dfrac{-\dfrac{13}{2}}{\dfrac{5}{2}}=-\dfrac{13}{2}\times\dfrac{2}{5}=\dfrac{13}{5}\)
Since \(m=\dfrac{-5-b}{2}\)...
I have to go, I guess I made a mistake somewhere! Good luck!

- theEric

\(b=\dfrac{-13}{5}\), to correct what I have above. Then \(m=\dfrac{-5-\dfrac{-13}{5}}{2}\\~\\\quad=\dfrac{-5+\dfrac{13}{5}}{2}\\~\\\quad =\dfrac{\dfrac{-12}{5}}{2}\\~\\\quad =\dfrac{-12}{10}\\~\\\quad=\dfrac{-6}{5}\)
Since \(y=mx+b\), \(y=\dfrac{-6}{5}x-\dfrac{13}{5}\).
I agree with @RadEn and now wonder where I went wrong with the other method! Thank you!

- theEric

Ah, in the very beginning, I made a mistake!
\(m=\dfrac{(-5)-1}{2-(-3)}\neq \dfrac{-6}{1}\)
\(m=\dfrac{(-5)-1}{2-(-3)}=\dfrac{-6}{5}\)
I believe I correctly solved for \(y\), and so my second flaw was just continuing to use my false \(m\)-value.
So \(\ y=mx+3m+1\\\quad=\dfrac{-6}{5}x+3\left(\dfrac{-6}{5}\right)+1\\\ \\\quad =\dfrac{-6}{5}x+\dfrac{-18}{5}+\dfrac{5}{5}\\\ \\\quad =\dfrac{-6}{5}x+\dfrac{-13}{5}\\\ \\\quad =\dfrac{-6}{5}x-\dfrac{13}{5}\)
\(y=\dfrac{-6}{5}x-\dfrac{13}{5}\quad\huge\checkmark\)
Now you see, @RadEn , @izhangy , and @PlsHaveMercyLilB , that the first method can in fact work! Not that \(2-(-3)\neq 1\), but \(2-(-3)=2+3=5\).
In conclusion, the method was correct, I just made an error in the calculation of \(m\) that was used in the calculation of \(y=f(x)\), so the \(m\) and \(b\) of the \(y=mx+b\) equation did not match their true values until the correction was made.

- theEric

Right, @PlsHaveMercyLilB . I chose to answer this question to prove that point. Your method was correct, and I proved that. RadEn and I also proved that it is sufficient to create and solve a system of equations. Both solutions came out to be the same.

- anonymous

ya thanks i didnt read the whole thing lol

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