Finding the equation of a line tangent to function f?

Question

Full Question: The line L is the tangent to the curve of f at (3,12). Find the equation of L in the form y = ax + b.

The original function given is f(x) = (1/3)x^3 + 2x^2 - 5x.

I tried to find the derivative by 'completing the squares' (which I dont fully understand) so it may be wrong. Either way, here is what I got.... it may be useful?

y = (1/3)x * (x+3)^2 - 7x

Thanks for help :]

justsomeguy · Answer

Completing the square is something different and isn't needed here.

Use this formula:

dy/dx= c*n*x^n-1

Where c is any constant on the term (in this case 1/3)
n is the power to which it is originally raised

So the derivative is x^2+4x-5. This formula will give you the SLOPE of the line tangent to the curve at the points given, (3,12).

So when x=3, the slope of the tangent is 3^2+4*3-5 which = 16.

Now you have a point (given as 3,12) and the slope, 16, so you should be able to plug that into the point slope formula and convert it into slope intercept form.

cantarero · Answer

a) f'(x) = 2x + 3 b) At (a million, 4), gradient = 2(a million) +3 = 5 So equation is y = 5x + c Substituting x and y values (a million, 4) we get 4= 5(a million) + c So c = 4 - 5 = -a million So the equation to the tangent is y = 5x - a million

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Finding the equation of a line tangent to function f?

2 Answers