1. ## Re: gradient at following points at the given curve?

The problem I think is that you don't understand the meaning of the derivative enough to understand the exercice. So, you have given a function $f(x)=3x^2-4x+1$ and you have to find the gradient at the point $x=1$ which you have found: 2 and this is the gradient of the tangent line at the point $x=1$, the only thing you need now is the point which is (1,0) which lies on the tangent line.

So this means you just have to substitute these values in the equation of a straight line:
$y-y_1=m(x-x_1)$
where m=gradient, and $(x_1,y_1)$ a point on the line.

2. ## Re: gradient at following points at the given curve?

c=-2 so overall y=2x-2?

3. ## Re: gradient at following points at the given curve?

Originally Posted by andyboy179
c=-2 so overall y=2x-2?
Yup

4. ## Re: gradient at following points at the given curve?

Originally Posted by andyboy179
im not sure how to do this next step. i can do the equation of a straight line but i only have one set of brackets which is (1,0) what do i do?
I say this with no ill-will: Unless you go back and thoroughly review your PRE-calculus, the world of calculus is going to be a world of pain for you. A lack of competency in basic algebra and Pre-calculus is the number one reason for students struggling with calculus.

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