anonymous
  • anonymous
Let f be a linear function where f(2)= -5 and f(-3)=1. Find f(x). HOW TO DO?!?!?!?!?!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
RadEn
  • 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
  • RadEn
got it ?
anonymous
  • anonymous
I still do not understand sir.

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anonymous
  • anonymous
They gave u two points that are on the line, so you can calculate the slope
anonymous
  • anonymous
do u know how?
anonymous
  • anonymous
what are the points..?
anonymous
  • 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
  • anonymous
okay and the other point is (-3,1), right?
anonymous
  • anonymous
yep
anonymous
  • anonymous
u know the formula for slope?
anonymous
  • anonymous
You mean the point slope formula?
anonymous
  • anonymous
You're gonna need the slope to use that formula
anonymous
  • anonymous
have u ever heard of rise over run?
anonymous
  • anonymous
ohh i know is it (y-y1)/( x-x1)?
anonymous
  • 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
  • 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
  • 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
  • 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
  • RadEn
got *
anonymous
  • anonymous
Ohh I see, so I use elimination method?
RadEn
  • RadEn
yeah, hold on :)
anonymous
  • anonymous
a=1 b=0
anonymous
  • anonymous
I cancelled out b.
RadEn
  • RadEn
subtract direct both equations, you will get : -2a + b = -5 -3a + b = 1 -------------- (-) a = -6
anonymous
  • anonymous
ohhh. so what to do after that
RadEn
  • 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
  • anonymous
alrighty, but the question ask to find f(x).
RadEn
  • RadEn
now, plug the values of a and b to f(x) |dw:1377654643235:dw|
anonymous
  • anonymous
oh so when it says find f(x), it's asking for an equation?
RadEn
  • 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
  • anonymous
what...
anonymous
  • anonymous
the answer is f(x)=-6x+13, right!?!?!?
RadEn
  • RadEn
wait. looks i make a mistake above
anonymous
  • anonymous
What mistake did you make?
RadEn
  • 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
  • anonymous
alright so, FINAL ANSWER is f(x)=-6x-11. RIGHT?!?!
RadEn
  • RadEn
my mistake in -2 * -6 = -12, that should be 12. yes, it is f(x) = -6x - 11
anonymous
  • anonymous
alrighty thanks man.
RadEn
  • RadEn
you're welcome :)
RadEn
  • RadEn
wait. the totally i mstake in calculation again above, so sorry
RadEn
  • RadEn
the first equation should f(2) = -5, a(2) + b = -5 or 2a + b = -5 ........ (1)
RadEn
  • RadEn
the 2nd equ already be correct f(-3) = 1, a(-3) + b = 1 or -3a + b = 1 ........ (2)
anonymous
  • anonymous
what does ......(1) mean?
theEric
  • 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
  • theEric
Does my work look good to you all? Good work, everyone! :)
anonymous
  • anonymous
wait so you got f(x)= 3x+19?
theEric
  • 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
  • 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
  • RadEn
nah, we got the different of the result @theEric . hahaha..
theEric
  • theEric
I see that! Haha!
RadEn
  • RadEn
just check, is it be right that f(-2) = -5 and f(-3) = 1
anonymous
  • anonymous
ok so who is right?
theEric
  • 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
  • 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
  • anonymous
K MAN, i believe you.
anonymous
  • anonymous
thanks all.
RadEn
  • RadEn
@theEric , do you agree with me, hehe :)
anonymous
  • 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
  • RadEn
|dw:1377657489398:dw|
anonymous
  • anonymous
ok
theEric
  • 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
  • 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
  • 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
  • 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
  • anonymous
ya thanks i didnt read the whole thing lol

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