This Blog exists for the collective benefit of all algebra students. While the posts are specific to Mr. Chamberlain's class, any and all "algebra-ticians" are welcome. The more specific your question (including your own attempts to answer it) the better.
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Wednesday, January 18, 2012
hw #5-4 Point-Slope Form
... Wanna remember the formulas??? Write 'em down every time you need to use 'em...
I should not have given you problem #31... it involves a concept called composite functions which we did not cover in class... you can kinda/sorta figure it out, but I can tell you WITH CONFIDENCE that it will not be on the test. I also guarantee you will see this concept in Algebra 2, so you have something to look forward to!!
Just for chuckles, ok, here goes: a) f(x)=2x-5 j(x)=f(x)+3 It's kinda simple, just replace f(x) with 2x-5 j(x)=(2x-5)+3 j(x)=2x-2
b) f(x)=2x-5 k(x)=f(x+3) this time, you are replacing the x in the first equation with x+3 SO... k(x)=f(x+3)=2(x+3)-5 k(x)=2x+6-5 k(x)=2x+1
... was that fun??? REMEMBER... NOT ON TEST (until Algebra 2!!)
Hi mr. c I am having trouble with number seven. I heard you go over something like it in class, and how the slope intercept and point slope forms are similar. I was wondering if you could go over that again. And was also wondering if there was a way to find the y-intercept without using a graph, probably a dumb question
We learned that an equation in point-slope form can be converted into an EQUIVALENT equation in slope-intercept form. As we have found (see the back wall of our classroom), equations in slope intercept are very useful, since we can easily graph the equation given the y-intercept (makes for a great starting point) and the slope (our stair-stepper).
ex. number 19. It says given these two points, (1,4) and (-1,1), write a an equation in point slop form for the line that passes through them. Then turn it into slope intercept form. How do you turn one into the other if you don't have the y-intercept to make the equation true. I have y-4=3/2(x-1) then i tried to turn it into slope intercept form. I have y=3/2x+? how do i find the y-intercept? What am i not getting?
NEVERMIND I GOT IT!!! you just solve for b in the slope intercept equation. Plug in the x and y coordinates and solve for b. Then you can finish the equation. So for 19 it would be y-4=3/2(x-1) for pointslope form. And for slope intercept it would be y=3/2x+2.5!!!! Ur so helpful mr. c you inspire confidence!!!!
What i figured out about the relationship b/w point slope form and slope intercept is that you basically simplify the point slope equation. Once you simplfy and get y by itself you have the equation for slope intercept (if im right)!! It helps a lot if you think that way!! :) Also problem 29 was very confuzzling. I didn't even know where to start writing the equation. Help?!
In #27, you are told the relationship is linear and you are given two points. We are asked to find the Celsius temp given the Fahrenheit temp, so we will say that the Celsius temp depends on the Fahrenheit temp, SO, C is the dependent and F is the independent variable.
Therefore, we have the points (50,10) and (77,25), agreed?
Can you calculate the slope? SURE!!!
The slope is 15/27, which simplifies to 5/9, yes?
NOW, we can use point-slope form to derive slope-intercept form, yes?
I should not have given you problem #31... it involves a concept called composite functions which we did not cover in class... you can kinda/sorta figure it out, but I can tell you WITH CONFIDENCE that it will not be on the test. I also guarantee you will see this concept in Algebra 2, so you have something to look forward to!!
ReplyDeleteJust for chuckles, ok, here goes:
a)
f(x)=2x-5
j(x)=f(x)+3
It's kinda simple, just replace f(x) with 2x-5
j(x)=(2x-5)+3
j(x)=2x-2
b)
f(x)=2x-5
k(x)=f(x+3)
this time, you are replacing the x in the first equation with x+3
SO...
k(x)=f(x+3)=2(x+3)-5
k(x)=2x+6-5
k(x)=2x+1
... was that fun??? REMEMBER... NOT ON TEST (until Algebra 2!!)
Hi mr. c
ReplyDeleteI am having trouble with number seven. I heard you go over something like it in class, and how the slope intercept and point slope forms are similar. I was wondering if you could go over that again. And was also wondering if there was a way to find the y-intercept without using a graph, probably a dumb question
Thanx
We learned that an equation in point-slope form can be converted into an EQUIVALENT equation in slope-intercept form. As we have found (see the back wall of our classroom), equations in slope intercept are very useful, since we can easily graph the equation given the y-intercept (makes for a great starting point) and the slope (our stair-stepper).
DeleteCapeesh?
ex. number 19. It says given these two points, (1,4) and
ReplyDelete(-1,1), write a an equation in point slop form for the line that passes through them. Then turn it into slope intercept form. How do you turn one into the other if you don't have the y-intercept to make the equation true. I have y-4=3/2(x-1)
then i tried to turn it into slope intercept form. I have
y=3/2x+? how do i find the y-intercept? What am i not getting?
NEVERMIND I GOT IT!!!
ReplyDeleteyou just solve for b in the slope intercept equation. Plug in the x and y coordinates and solve for b. Then you can finish the equation. So for 19 it would be y-4=3/2(x-1) for pointslope form. And for slope intercept it would be
y=3/2x+2.5!!!!
Ur so helpful mr. c you inspire confidence!!!!
... some might say helpful... I prefer to say "brilliant"
ReplyDeleteGosh, I really hope EVERYONE is at least reading the blog... ANYBODY THERE??? HELLO-OOOOO!!!!
ReplyDeleteWhat i figured out about the relationship b/w point slope form and slope intercept is that you basically simplify the point slope equation. Once you simplfy and get y by itself you have the equation for slope intercept (if im right)!! It helps a lot if you think that way!! :)
ReplyDeleteAlso problem 29 was very confuzzling. I didn't even know where to start writing the equation. Help?!
I'm struggling with pg 318 #8-11 can you remind me how to solve them?
ReplyDeletesome of you asked about #27...
ReplyDeleteIn #27, you are told the relationship is linear and you are given two points. We are asked to find the Celsius temp given the Fahrenheit temp, so we will say that the Celsius temp depends on the Fahrenheit temp, SO, C is the dependent and F is the independent variable.
Therefore, we have the points (50,10) and (77,25), agreed?
Can you calculate the slope? SURE!!!
The slope is 15/27, which simplifies to 5/9, yes?
NOW, we can use point-slope form to derive slope-intercept form, yes?
So:
y-y1=m(x-x1)
y-10=(5/9)(x-50)
y-10=(5/9)x - (250/9)
y-(90/9)=(5/9)x - (250/9)
y=f(x)=(5/9)x - (160/9)
Slope is 5/9 and y-intercept = -160/9
So, now we can substitute ANY Fahrenheit temp and obtain a Celsius temp.
Gosh, we end up with an f of F... how about that??
C=f(F)=(5/9)(59) - 160/9
C = 295/9 - 160/9
C = 135/9 (how many times does 9 go into 135??)
C = 15
So, when the temp is 59F it is 15C... Brrrrrrr!!
Capeesh?