Problem 1: Find an equation for the line with y-intercept 3 that is perpendicular to the line y=2/3x-4. & Problem 2: If P(4,-5) is a point on the graph of the function y=f(x), find the corresponding point on the graph of y=2f(x-6).
2006-11-23 10:31:33 · 10 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Problem 1: Find an equation for the line with y-intercept 3 that is perpendicular to the line y=2/3x-4
to find a line perpendicular to another line, use the opposite reciprocal of the slope
in this case, the slope is 2/3, therefore the opposite reciprocal of the slop is -3/2
now you know the slope and you also know that the y-intercept is 3
use the equation y = mx + b
if m is the slope and b is the y-intercept, then your equation is
y = (-3/2)x + 3
-------------------------------------------
Problem 2: If P(4,-5) is a point on the graph of the function y=f(x), find the corresponding point on the graph of y=2f(x-6)
y = 2f(x-6)
y/2 = f(x-6)
y = -5
x = 4
plug them into the equation
-5/2 = f(4-6)
-5/2 = f(-2)
x = -2
y = -5/2
thus the answer is
x = -2
y = -5/2
having trouble with this one myself... sry... dont know the answer
2006-11-23 10:53:24 · answer #1 · answered by trackstarr59 3 · 0⤊ 1⤋
The equation of a linear line is in the form of y=mx+b. m represents the slope of the line and b represents the y-intercept. well since it is given that the y-intercept is 3 we have the partial equation y=mx+3. so what we need to know is the slope of the line. if the line is perpendicular to the line y=2/3x4, this means that the slope of the line will be the negative reciprical of 2/3 which is -3/2. so the final equation we get is y=-3/2x+3. fairly simple, once you get the hang of it. now let's take a look at question number 2.
P(4,-5) is a point on the graph of y=f(x) then the corresponding point on the graph is :
y=2f(x-6) ignore the 2 for now. so we gwt
y=f(x-6)
plug in 4 and we get y=f(4-6) ==> y=f(-2) y is equal to -2
we can't just take out the two it'll totally change the problem so take -2 and divide it by 2 that means y=-1 so the corresponding point is (4,-1)
2006-11-23 18:57:55 · answer #2 · answered by Carpe Diem (Seize The Day) 6 · 0⤊ 1⤋
Problem 1:
Find an equation for the line with y-intercept 3 that is perpendicular to the line y=2/3 x - 4.
Slope of line is 2/3 so slope of perpendicular line is -3/2 (as 2/3 *( -3/2) = -1)
Thus line is of form y = -3/2 x + b
Y - intercept is 3 therefore b = 3
So required line is y = -3/2 x + 3
OR, in general form, 3x + 2y - 6 = 0
Problem 2:
If P(4,-5) is a point on the graph of the function y = f(x), find the corresponding point on the graph of
y=2*f(x-6) ie y/2 = f(x - 6) ie Y = f(X) where X = x - 6 and Y = y/2
So (4, -5) â (-2, -5/2)
2006-11-23 18:42:18 · answer #3 · answered by Wal C 6 · 0⤊ 2⤋
Problem 1: Find an equation for the line with y-intercept 3 that is perpendicular to the line y=2/3x-4.
slope perpindicular to a line with slope 2/3 is -3/2
y-int=3
y=-3x/2+3
Problem 2: If P(4,-5) is a point on the graph of the function y=f(x), find the corresponding point on the graph of y=2f(x-6).
(4,-10)
2006-11-23 18:34:07 · answer #4 · answered by yupchagee 7 · 1⤊ 2⤋
Wow! A REAL question!
I can't answer this one of the top of my 3 byte brain, but I can relate to your dilema -- I failed algebra more than once in Highschool
However I eventually got very good at algebra, calculas and trig. AFTER I joined the Airforce Reserve and became an Avionics Electronics specialist.
No I didn't get any smarter, but the big differance is that instead of studying just the math, I was now seeing how the math appied to real life objects and situations. -- This made all the differance for me. I even got to the point where I was begining to understand fourrier transforms and Antenna and wave guide mathematics which is some really hairy stuff.
So don't look to this "dear abby" webpage for a quick answer. That is a cop out -- If I would have had the Internet availible to me when I was growing up, I don't think I would have failed ANY math classes in school.
First decide what field of work you most want to do (because all most all of the good ones use math -- sorry)
get your search page up and running, and go find out where they are using the polar coordinate algebra you are trying to solve now, and then see if the "real world use" of the same type of math improves your understanding.
Don't give up, because if you do there are about a thousand kids from other third world countries that are working on the same problems, and they are motivated -- they want to be where you live now, and they will be happy to take the job that you want to apply for later.
(I failed English too as you can probably tell by my spelling and punctuation.) :) -- don't give up!
2006-11-23 19:02:30 · answer #5 · answered by nonlinear_systems 2 · 0⤊ 1⤋
Answer to problem 1.
The line perpendicular to
y=(2/3)x - 4
having its y-intercept at +3 is
y=(-3/2)x + 3
You were given the y intercept. All you had to know was that the slope of a perpendicular line is the negative reciprocal of the slope of the original line.
2006-11-23 19:08:47 · answer #6 · answered by Anonymous · 0⤊ 1⤋
Problem 1: y= 2/3x+3
therefore: m1=2/3 so m2=-3/2
now
using m2=-3/2 and y intercept=3
the perpendicular equation ,,,, y= - 3/2+3
sorry but i don't really understan problem 2
2006-11-23 18:46:33 · answer #7 · answered by Nilesh R 1 · 0⤊ 1⤋
Here is the answer to the first question but I have to run and can't finish the second.....
y=3/2x+3
2006-11-23 18:35:15 · answer #8 · answered by Anonymous · 1⤊ 3⤋
All the birds fly south for winter.
That maths question was easy.
2006-11-23 18:33:22 · answer #9 · answered by Anonymous · 0⤊ 4⤋
CHEATER
2006-11-23 18:39:35 · answer #10 · answered by RED-CHROME 6 · 0⤊ 4⤋