# Thread: Find equations of the tangent lines at points

1. ## Find equations of the tangent lines at points

Given y = 3 + 4x^2 - 2x^3, how do I find the slope of the tangent line at point (1,5), for example, using the Newton Quotient Law or something in the Derivatives Chapter I'm studying. I "cheated" and estimated the slope by plugging in x = 1.01 and then finding the slope of between the points (1,5) and (1.01, 5.019798) but I'd like to be able to solve this the proper (and required) way. My work is attached.

Any help would be greatly appreciated!

2. $y = 3 + 4x^2 - 2x^3$

$y' = 8x - 6x^2$

$when\ x=1\ m= 8(1)-6(1)=2$

3. How did you go from 3 + 4x^2 - 2x^3 to 8x - 6x^2 ?

4. You differentiate it.

5. Could you please be a little more precise because I am very knew to this?

6. differentiation. the rule is:

If $y = x^n$ then y' = $nx^{n-1}$ so if $y= x^2$ then the differentiated form will be $2x$
The derivative of any constant. e.g. 9, 8, 7, 100, is zero.
The derivative of an equation of the form $y = ax^n$ is $y' = anx^{n-1}$

so for this equation 3 + 4x^2 - 2x^3

the derivative of 3 is zero
the derivatie of $4x^2$ is $8x$
the derivative of $-2x^3$ is $-6x^2$

Put it all together and the derivative is:
$8x - 6x^2$

7. Ok thanks, I get how to do the differentiation thing now but why am I doing that? Like what I am finding? (in not so mathematical terms please)

8. You're finding the gradient function. This is the function that allows you to find the gradient. Once you have the gradient function, any value substitituted in this, will determine the gradient of a certain value at the point substituted.

e.g. $y = x^2$

dy/dx = 2x

substitute x =1 into dy/dx.
this is equal to 2.
therefore, the gradient when x =1, is m =2

so basically, when you differentiate something, you are given another equation known as the GRADIENT FUNCTION. This is an equation specifically targeted at finding gradients.