Find an equation of the tangent line to the curve at the given point.?

Question

a) y= 2x^(3) - 5x , (-1,3)
b) y= 8x/(x+1)^2 , (0,0)

and please explain...

Answer 1

The equation of a line is
[1] ... y = mx + b
where m is the slope.

The slope of a tangent in a point (x,y) is equal to the derivative of the function in x: m = y'(x). When you know m you can substitute the known values (x, y) in equation [1] and solve for the intercept b.

a) y(x) = 2 x^3 - 5 x
--> y'(x) = 6 x^2 - 5
--> y'(-1) = 6 (-1)^2 - 5 = 6 - 5 = 1

So the equation for the tangent line is
[2a] ... y = 1 x + b
Substitute the point (-1, 3):
[2b] ... 3 = -1 + b
therefore b = 4 and the solution is
[2c] ... y = x + 4

b) y(x) = 8 x (x+1)^(-2)
--> y'(x) = 8 (x+1)^(-2) + 16 x (x+1)^(-3)
[product rule!]
--> y'(x) = [8 (x+1) + 16 x] / (x+1)^3
--> y'(0) = 8

So the equation for the tangent line is
[3a] ... y = 8 x + b
but because the line goes through the origin, b = 0 and the solution is
[3b] ... y = 8x

Answer 2

take the derivative of the function and substitute for x to get the slope of the tangent then using slope-point formula to find the equation so:
a)
y'=6x^2-5
=6(-1)^2-5
=1

y2-y1=m(x2-x1)
y2-3=1(x2+1)
y=1-x+3
y=x+4

b)y'=[8(x+1)^2-16x(x+1)]/(x+1)^4
substitute 0 for x
y'=8

then just substitute into y2-y1=m(x2-x1)