fanficgirl:
Please can someone help me. I've forgetten how to do all these math problems and I really need assistance and google isn't helping. Would really appreciate it...
1)Write the equation of a line of a line in point slope form that passes through the point (5,-2) and is perpendicular to 3x-2y=2

GMacbeth:
2y=3x-2
y=3x/2-2/2
gradient=3/2. also gradient y-5/x-2
for perpendicular gradient m1xm2=-1
(y-5)/(x-2) x 3/2 =-1
3y-15/2x-4=-1
3y-15=-2x+4
3y+2x=19

GMacbeth:
2y=3x-2
y=3x/2-2/2
gradient=3/2. also gradient y-5/x-2
for perpendicular gradient m1xm2=-1
(y-5)/(x-2) x 3/2 =-1
3y-15/2x-4=-1
3y-15=-2x+4
3y+2x=19

fanficgirl:
Aww men I did it wrong then D:

shaboti:
he z wrong

shaboti:
he z wrong

fanficgirl:
How do you do it right then? Yeah, isn't his gradient wrong? Sorry i'm mathmatically stupid...

shaboti:
I solved with a slightly diff method tho.

dy/dx of 3x-2=3 is 3/2 (gradient)

But for a line perpendicular to it, the gradient will be the negative inverse.

Therefore 3/2 will change to -2/3.

To find the eqn of the line, u can use this method
Y-y' = m (X-x').. where m represents the gradient.

Substututing..

Y+2=-2/3 (X-5)
Y+2=-2X/3 + 10/3
3Y+6 =-2X+10
3Y+2X=10-6

3Y+2X=4

shaboti:
I solved with a slightly diff method tho.

dy/dx of 3x-2=3 is 3/2 (gradient)

But for a line perpendicular to it, the gradient will be the negative inverse.

Therefore 3/2 will change to -2/3.

To find the eqn of the line, u can use this method
Y-y' = m (X-x').. where m represents the gradient.

Substituting..

Y+2=-2/3 (X-5)
Y+2=-2X/3 + 10/3
3Y+6 =-2X+10
3Y+2X=10-6

3Y+2X=4

fanficgirl:
Really sorry to bother you, and hope i'm not being really annoying but to graph the equaton y=-x^2(x+3)((x-1))^2 what would you do with the negative x^-2 squared? Or wold the graph start negative and then would go to -3 then slope down and then up to zero and then to 1 where it would end down again right?

shaboti:
I dont understand you question.. what are your range of values , and why did you add double brackets to x-1?

fanficgirl:
There isn't any range of values, thats the question and its just supposed to be a simple sketched graph. Sorry I meant to say that (x-1) is squared too with the brackets, should have done it better.

shaboti:
hmm.. as far as I know there shud be values to work with, or are you expected to just use your own values?
Perhaps this is beyond me.

fanficgirl:
This is an example...

shaboti:
oh I see what you mean now, and yes I believe it will run through -3, through the origin nd den to +1

(-x)^2 =0
X=0

fanficgirl:
Please can someone help me. I've forgetten how to do all these math problems and I really need assistance and google isn't helping. Would really appreciate it...
1)Write the equation of a line of a line in point slope form that passes through the point (5,-2) and is perpendicular to 3x-2y=2

Umartins1:


3x-2y =2

-2y= 2-3x

y= -2/2 + 3x/2

M₁= 3/2


When two lines are perpendicular,

M₁M₂= -1

M₂= -1/M₁

M₂= -1/ 3/2

M₂ = -2/3


Equation of a line=

y-y₁= m (x-x₁)

y-(-2)= -2/3 (x-5)

y+2 = -2/3 (x-5)

3(y+2) = -2(x-5)

3y+6= -2x+10

3y+2x+6-10=0

3y+2x-4=0

fanficgirl:
Thanks! Yeah someone already helped me figure it out but glad its right.

Umartins1:


The first person was wrong.

fanficgirl:
Oh good so then its correct, thank you so much! But how would the graph be like if its an exponent like -5 ^(x-1)+3

shaboti:

could not arrive at anything miss..
5^(x-1)+3 =5^x * 5^-1 * 5^3
5^x=0
x=0?

Dunno biko
Btw what class are u in?

fanficgirl:
Calc, were supposed to be doing reviews from other years and its embarrassing that I can't remember how to do 90 percent of it... And oh its just supposed to be graphed not solved..

fanficgirl:
Oh, I thought it didn't matter which way it was simplified. So your simplified version is the definite correct way then?

shaboti:
hm. Secondary sch?

Umartins1:


Check his final answer. My answer and the one of the guys that used differentiation are both correct.

fanficgirl:
Yes

fanficgirl:
Oh good because that was the one I was referring to. Thanks!

shaboti:
ss3 i guess?