Question

Medal: if f(x)=x^3-5x+1, find f'(1) okay, I know how to do that, it's easy. What they then ask me to do is where I get lost:

Use it to find an equation of the tangent line to the parabola y=x^3-5x+1 at the point (1,-3). y = ?

Answer 1

f(x)=y

Answer 2

You said that you have the derivative value, f'(1). Remember what that is - it's the SLOPE of the GRAPH at the point x=1. Hence, it is also the SLOPE of the TANGENT LINE to the graph at x=1.

So.... you have the slope (that is the value you computed, f'(1)) of the line.

You have a point ON the line: (1, -3)

Given slope and a point on the line, do you know how to find the equation for a line?

Answer 3

y=mx+b? I'm not 100% sure.

Answer 4

if you derive the function, you get a general equation for slope of the function, where you can evaluate the slope of the original function for every x.

now they want the tangent line at some position. we already know the slope: it will be f'(position). also, like Sweetangel14 said, the position of the tangent line will be so that it touches the function at the position,.....

Answer 5

OK, that's the slope-intercept form of the equation for a line.

So you have m, right? That's your slope, which is just f'(1).

Now all you need is b. You could go a couple of ways here:

1. use the given point and the slope that you already have, plug into y=mx + b and solve for b.

2. Use the given point and the slope that you already have, and the point-slope equation for a line:
\(\large y-y_1=m(x-x_1)\)

These methods should be familiar to you from your algebra or pre-calculus math classes.