How do I find the tangent to the curve y = x^3 - 4x + 1 where x = a?

Question

I assume I need to use the formula limx->a (f(x) - f(a)) / (x - a) but I'm not sure... any suggestions?

The answer is 3a^2 - 4 according to my book. But I have no idea how to get there.

I still don't get how to arrive at the derivative... you all seem to be jumping there instantly. How do I get the derivative (the answer).

Answer 1

y = x^3 - 4x + 1

To find the slope of the tangent line at x = a, the first step is to take the derivative.

y' = 3x^2 - 4

Now, the slope is determined to x = a, so we plug in a for every instance of x, and

m = 3a^2 - 4

If we wanted to go a step further and find the equation of the tangent line at x = a, then we use the slope formula for
(a, a^3 - 4a + 1) and (x, y), replacing (x1, y1) and (x2, y2), for the formula

(y2 - y1) / (x2 - x1) = m
(y - (a^3 - 4a + 1)) / (x - a) = 3a^2 - 4

y - (a^3 - 4a + 1) = (3a^2 - 4)(x - a)
y = (3a^2 - 4)(x - a) + a^3 - 4a + 1

And so forth.

Answer 2

the tangent is the derivate of the function at x=a
y'=3x^2-4
tg(alfa)=3*a^2-4

Answer 3

a function has a tangent when y'=0
get the idea?
differentiate the function and solve for x=0 what u get are points of tangetnts

Answer 4

f(x) = x³ - 4x + 1
f `(x) = 3x² - 4
f `(a) = 3a² - 4 and this is the gradient of the tangent to the curve
when x = a.
Is this what you mean when you say ---"find the tangent" ?

Answer 5

where x=a y=a^3-4a+1
dy/dx=3x^2-4
at x=a dy/dx=3a^2-4
so the equation
y-(a^3-4a+1)=(3a^2-4)(x-a)
y-a^3+4a-1=3a^2x-3a^3-4x+4a
3a^2x-y-2a^3+1=0

Answer 6

mmm....