derivatives- slopes and tangents

G

Guest

Guest
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
 
I believe this will work: Plug in the given point. This gives you one equation in a, b, and c. Plug in the origin. This gives you another equation in a, b, and c. Differentiate and note that dy/dx = 1 at the origin. This will give you a third equation. Solve the system.

Eliz.
 
can u pls explain that one more time.. i plugged in 1 for x so i ogt y=a+b+c then i plugged in 0 for x and got y=c.... am i doing this right?
 
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

y=ax^2+bx+c sub in the points (1,2)
2 = a +b +c


dy/dx= 2ax +b

tangent to the line y=x, this has a gradient of 1 (y=mx+ b)
also the x and y values are the same y=x
1=2ax +b

ok do you want to have a go.......
 
ok --back again.

y=ax^2 + bx + c

put in (1,2)
2= a + b + c

put in (0,0)
0 =0 + 0 + c

therefor c= 0

next at the dy/dx=1 (as per last post)
dy/dx= 2 ax + b
1= 2 ax + b
now sub in the (0,0) point ie x=0
1= b
now solve a
 
43) find the tangents at the origin and the point (1,2) of y=4x/(x^2+1)
 
Please review the replies posted to the other thread(s) you've started on question 43. Thank you.

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

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

y = u . v where u =4x and v = (x^2 +1)^-1

du/dx= 4

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

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

sub the above parts in and show me how you go....
 
i dont understand how u got that derivative.. i think i did the derivative wrong... can u explain it 1 more time pls
 
i dnt understand y u separated the original equation... like when u broke it up into 2 parts the u and the v why cant u just take the derivative of the whole thing
 
this type of equation consists of 2 seperate statements of "x" which are multiplied together and this is the solution method.
 
You must log in or register to reply here.
Top