Question

how do you find the equation of the tangent line to the curve 2x^2-2x+1 that has a slope of 2?

Answer 1

So the first thing we need to do is differentiate the curve, in order to figure out the general slope of the tangent line. There are a couple of ways of doing this, but we're going to use the quick way, which relies on two rules.

Firstly, the derivative of x^n is nx^n-1. This also extends to multiples of x. cx^n = cnx^n-1. (So 3x^n = 3(nx^n-1).)

Example: 2x^3. c = 2, n = 3. The derivative is thus 6x^2.

The second rule is that (d/dx u) + (d/dx v) = d/dx (u + v). E.g: the derivative of x^3 plus the derivative of x^2 = the derivative of (x^3 + x^2). This allows us to differentiate any polynomial by applying the first rule to each of it's terms, then adding together the results. (This even works on negative exponents.)

So let's do that. d/dx 2x^2-2x+1 = (d/dx 2x^2) + (d/dx -2x) + (d/dx 1)

By our two rules earlier, that equals 4x - 2 + 0, or 4x - 2. (The derivative of a constant c is always 0, because in a graph such as "x = 3" the slope of that line is always 0. Draw it and see if you're unsure.)

Technically, by the wording of your question, we're now done. But if the slope is 4x-2, how can the slope be 2? That's because 4x-2 is the general slope of the tangent line, and we can now plug in values of x to get the slope for a specific point on the curve. In this case, simple algebra will determine that the slope = 2 when x = 1:

4x - 2 = 2
4x = 4
x = 1

And we're done! The equation is 4x-2, and this has a slope of 2 when x is equal to 1.

Answer 2

@Salivanth
to complete the problem, they want the equation of the tangent line.
As you found, the tangent line to the curve at the point (1,1) will have a slope of 2
and its equation will be
y-1 = 2(x-1) or
y = 2x -1

A plot is attached.