derivatives: slopes and tangents

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41) a)find an equation for the line perpendicular to the tangent to the curve y=x^3 - 4x+1 at the point (2,1)
b)find equations for the tangents to the curve at the points where the slope of the curve is 8

43) find the tangents at the origin and the point (1,2) of y=4x/x^2+1

45) the curve y=ax^2+bx+c passes thru the point (1,2) and is tangent to the line y=x at the origin... find a,b,c
 
a)find an equation for the line perpendicular to the tangent to the curve y=x^3 - 4x+1 at the point (2,1)

tanget to the curve means find the gradient at this point
dy/dx = 3x^2 -4
sub in the given points to find dy/dx or the gradient

next a perpendicular line has gradients that multiply to give -1
ie m1 x m2 = -1

with this final graident and the original points sub them into y=mx + b
and you have the equation
 
can u pls explain when u say sub in the given points into dy/dx??
do u plug in (2,1) to y=3x^2-4?

ok i got the perpendicular slope of -1/8 now how do i find the y inetcept?
 
for part b)
make the dy/dx =8
and solve for the x point.
Use this x point to find the y point (use the original equation)
and repeat the y=mx +b process using m=8, and the new x,y points.

find the tangents at the origin and the point (1,2) of y=4x/x^2+1

do the dy/dx step again for this equation and sub x= 0 and x= 1 into this
 
can u pls explain when u say sub in the given points into dy/dx??
do u plug in (2,1) to y=3x^2-4?

plug in the x value and it will give you the dy/dx value (the gradient)

note it should be written as this dt/dx= 3x^2 -4, not y=......
 
ok i plugged in the x value and got 8 so the perpendicular slope is -1/8
but how do i find the y intercept?
 
ok you have your gradient of -1/8 and a set of points it goes through(2,1)

use this in y=mx + b to find b

then rewrite this equation using your m and b values ---you have the tangent equation.
 
wow u def r amazing.. so much better than my calc teacher!!! :D
 
Thanks for the feedback...keep going on the next ones I will be on line for a while.
 
umm ok for #43 what answer should i get cuz i got a dif answer than the back of the book... i took the deriv of the equation and got -4/x^2+1... so then i substituted x=0 and x=2 and i dnt think i got the right anwser
 
actually i got one of the answers when i substituted x for 1 but i didnt get it when i substituted x for 0
 
find the tangents at the origin and the point (1,2) of y=4x/x^2+1

can you clear this equation up for me as to which format it is

y=4x/(x^2+1) or

y=(4x/x^2 ) +1

then we can work on it...
 
i gtg eat dinner but is it ok if i cum back on in like 45 mins?? r u still gonna b here?
 
y=4x/(x^2+1)

ok need to find the dy/dx form of the above and this will be an long exchange....

y= (4x) (x^2+1)^-1

let u= 4x and v = (x^2+1)^-1

then dy/dx = u . dv/dx + v . du/dx

from above..
u= 4x
du/dx =4

v= (x^2+1)^-1
dv/dx = -2x (x^2+1)^-2

then dy/dx = u . dv/dx + v . du/dx
now replace these with the parts from above

This is the gradient equation..... sub in x=0 and x=1 to find the gradients

ok ---questions
 
thnk u soooo much i'll be on in a little i hope ur still here
 
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