To find the equation of a line tangent to a curve, do you use the definition of a derivative

for example, for : y=7x^3 - 2x^2 + 3x - 5 at x= -1
I use (f(x+h)-f(x))/h then I find the slope using the point slope formula? is that right?

Let y=f(x) = 7x^3 - 2x^2 + 3x - 5
f(-1) = -7 -2 -3 -5 = -17 >>>> (-1,-17)
y ' = 21x^2 -4x +3
y ' ( -1) = 21 +4 +3 = 28
To find equation of tangent substitute m = 28 & point ( -1,-17) into
y - y1 = m( x-x1)
y - - 17 = 28 ( x - -1 )
y +17 = 28 x +28
y = 28x + 11

Yes, but, if you haven't been, you need to take the limit as h --> 0.

With f(x) = 7x^3 - 2x^2 + 3x - 5, we see that the slope of the tangent line at x = -1 is:
f'(-1) = lim (h-->0) [f(-1 + h) - f(-1)]/h.

At this point, just plug and chug to get your slope and use point-slope form to get the equation of the line.

I hope this helps!

yes:
it will give you:
[7(x+h)^3-2(x+h)^2+3(x+h)-5-(7x^3-2x^2…
[7x^3+21x^2h+21xh^2+7h^3-2x^2-2xh-h^2+…
all x terms (without h terms) will cancel
[21x^2h+21xh^2+7h^3-2xh-h^2]/h
21x^2+21xh+7h^2-2x-h
as the limit of h approaches zero (plug in 0 for all h)
21x^2-2x
now plug in x=-1
21+2=23<---thats your slope
in original function find f(-1):
7-2-3-5=-3
the point of tangency is (-1,-3)--->(x1,y1)
use point slope form [y-y1=m(x-x1)]to find tangent line equation:
y+3=23(x+1)
then solve for y
y=23x+20

Note: d/dx ( x^n ) = n x^(n-1)

... y = 7x^3 - 2x^2 + 3x - 5
or [ @ x = -1 ] y = 7(-1)^3 - 2(-1)^2 + 3(-1) - 5 = -17
or (x,y) = (-1,-17)

dy/dx = slope = m = 21y^2 - 4x + 3 [ @ x = -1 ] = 21(-1)^2 - 4(-1) + 3 = 28

... y = mx + b
or y = 28x + b ← plug in slope
or -17 = 28(-1) + b ← plug in point
or b = 11

... y = 28x + 11 ← final solution in slope-intercept form

hmmmm...a food question. i like food questions. let's see...

seems like the first derivative of the equation of the curve is the equation of the tangent.

it's just that simple.

good luck.

use
f '(x) = 21x^2 - 4x + 3 ---> f ' (-1) = 28
and f(-1) = -17
---> the tangent equation y+17 = 28(x+1)