# Math Help - Tangent line derivatives

1. ## Tangent line derivatives

Hi,

I know this question is easy but I havent done this topic in awhile and I forget how to do it.

$f(x)=-0.1x^3 -0.26x^2 + 0.861x + 1.539$
I need to sketch the tangent lines to y=f(x) at x=-3 and x=2?
How do I work out the tangent line?

2. To find the tangent you need to take the derivative of f(x)

3. And when you found f'(x), you just replace x be the values given to you.

4. Hi, thats what I done, I got -0.279

So on my graph when x=-3, the corresponding y value is -0.279 ?

it looks like the tangent will cut through the curve?

5. No, the gradient at the x coordinate x = -3 is -0.279.

To find the y value, you need to replace in f(x).

But since you replaced in f'(x), you get the value of the gradient.

6. ok so when i plug in f(-0.279) I get a value of 1.28 for y.

so I connect from y=1.28 on the graph to where x=-3 line drops vertically onto the curve?

7. I'm getting f(-3) = -0.684. Check your calculations again.

You might want to find the equation of the gradient, by using:

$\dfrac{y - y_1}{x - x_1} = m$

You will get the y intercept of the tangent then.

Then, you connect the y intercept with the point to where the line x = -3 drops vertically on the curve.

8. You got f(-3) or f'(-3) = -0.684?

9. I didn't put any apostrophe in my answer, meaning I got f(-3) = -0.684. f'(-3) = -0.279.

You need to find the point on the curve first. You were given x = -3. The y coordinate is given by f(-3).

Now, since you are looking for a tangent at that point, you also need to find f'(3).

Once you get those, you have your coordinates.

f(-3) = -0.684
f'(-3) = -0.279
x = -3

Or, (-3, -0.684) with gradient m = -0.279

Thus, you get the equation of the tangent as being:

$\dfrac{y - (-0.684)}{x - (-3)} = -0.279$

This becomes:

$y = -0.279x - 1.521$

Now that you have the equation of the tangent, you can join the points: (-3, -0.684) and (0, -1.521) to get the tangent.