What is the equation of the tangent line to the graph of f(x) = x^3 - 4 at the point where x = 1 ?

Question

possible solutions...


y = 12x - 20

y = 12x + 12

y = 3x - 2

y = 3x - 6

Answer 1

f'(x)=3x^2
f'(1)=3
f(1)=-3
-3=3(1)+b
-6=b
y=-3x-6

Answer 2

What is the equation of the tangent line to the graph of f(x) = x^3 - 4 at the point where x = 1

dy/dx = 3x^2 which = 3 when x=1.
f(x) =y = - 3 when x =1
So y= mx+b = 3x +b
So -3 = 3+b --> b = -6
y= 3x -6 is the required equation

Answer 3

f(x) = x^3 - 4

We want to find the equation of the tangent line at x = 1. Note that at x = 1, y = f(1) = 1^3 - 4 = 1 - 4 = -3. So in essence we want to find the tangent line at the point (1, -3).

First, we obtain the slope of the tangent line by finding the derivative.

f'(x) = 3x^2. To find the slope at x = 1, we solve for f'(1).

f'(1) = 3(1)^2
f'(1) = 3

Therefore, our slope m = 3.

To find the equation of the line, we use the slope formula

(y2 - y1) / (x2 - x1) = m

And we use the point (1,-3) and (x, y) for (x1,y1) and (x2,y2). We also use m = 3.

(y - (-3)) / (x - 1) = 3

Solving this will get us the equation of the line.

y + 3 = 3(x - 1)
y + 3 = 3x - 3

y = 3x - 6, which should be the equation of the tangent line.