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f'(x)=2x f'(a)=2a so the equation of the line is y=2ax+b we know a point on this line (1,-3) so we -3=2a(1)+b -3=2a+b ....
One would have a negative slope, the other would be positive
Would those two lines be orthogonal to each other?
that looks possible it looks like it could form a 90 degree angles
Ok so then they must have reciprocal slopes
opposite reciprocal slopes
wait i htink i'm fixing to figure something out....
now we need to figure the y intercept for each line which is easy since we know a point on both lines. both lines have the point (1,-3)
y=6x+b (1,-3) -3=6(1)+b -3=6+b -3-6=b -9=b so y=6x-9 is the line that is tangent to the point (3,9) on the curver y=x^2 that passes through (1,-3) y=-2x+b (1,-3) -3=-2(1)+b -3+2=b -1=b so y=-2x-1 is the line tangent to the point (-1,1) on the curve y=x^2 that passes through (1,-3)
we can check if you want Let's find the tangent line at (3,9) and see if passes through (1,-3) so f'(x)=2x f'(3)=6 y=6x+b 9=6(3)+b 9-18=b b=-9 so y=6x-9 is (1,-3) on that line -3=6(1)-9=-3 so yes! now let's check the tangent line at (-1,1) and see if passes through (1,-3) so f'(x)=2x f'(-1)=-2 y=-2x+b 1=2+b b=-1 y=-2x+-1 is (1,-3) on that line -3=-2(1)+-1=-3 so yes! :) we are finished!
so the lines arent orthogonal
do you have any questions?
Why did you choose the general points (a,a^2) to be X1,Y1 respectively? Does it make a difference if I were to use them as X2,Y2 in the slope formula?
(a^2-(-3))/(a-1)=(-3-a^2)/(1-a) we can make both sides look the same (-1)/(-1)=1 so if we multiply our second fraction by (-1)/(-1) we would get the first fraction or vice versa
a-b is the same as -(b-a)
Yes thank you very much. I have just a few more questions if you don't mind helping me out with them.
go for it
If f(a) = 0 and f'(a) = 6, find the limit h --> 0 of: f(a + h) ------- 2h
i got it scanning it now
i just used the definition of derivative
and plugged and what they gave me and solved for limh->0 (f(a+h)/(2a))
do you understand both problems we did?
oops solved for limh->0(f(a+h)/(2h))
I don't understand how you did the second question.
what is definition of the derivative at a of f
Wait, ok wouldn't it be 12 for the answer because: 6 = f(a+h)/h * (1/2) 6(2) = f(a+h)/h
Sorry I don't know what happened
doesn't the question above ask for lim h->0 (f(a+h)/(2a))
oops the bottom is suppose to read 2h
so we have 6=limh->0(f(a+h)/(2h))
oops 6=limh->0(f(a+h)/h) do you understand this far?
ok but the want 1/2* limh->0(f(a+h)/h) so if we multiply the above equation on one side by 1/2 don't we have to do it to the other side?
Got it. Thank you.
Ok last question for now is this: Determine the value of "a', given that the line "ax - 4y + 21 = 0" is tangent to the graph y = a/x^2 at x = -2
if you write that tangent line in y=mx+b form we can find the slope of the line
so the equation of that line can be written y=ax/4-21/4 where a/4 is the slope of this tangent line
The slope of the line is a right?
ok so then we equate the two equations together
so you are talking about finding the derivative of that function y=a/x^2 at x=-2 and then setting that equal to the slope of tangent line and solving for a? that's what I'm working on right now
yes and I got 7 when equating the two equations together. But I don't understand how we can equate a y with a y' (derivative)
so we get a=a so this equation holds for any a is what I get
Oh so you are making them equal to each other by solving for a?
a = , a =, and then equating them?
derivative means slope this slope is suppose to be the same as the slope tangent to whatever point we are talkng about so we can find the slope of the tangent line and find the derivative of the curve at the given point and set them equal to find what a should be but the resulting equation is a=a this equation holds for every number 3=3 4=4 it is never not true
a/4=-2a/(x^3) at x=-2 is what I did
how did you get a/4?
a/4 is the slope of the tangent line
OOHHHH I understand So we can take a/4 to be the slope, which must be the same as the graph's slope at that point?
right just like the very very question we did sort of
Got it. Thank you so much for your help. I have to go now. I really appreciate you taking your time to do this for me.
no problem become my fan :)
Sure, see you.