Given a parabola $y= x^2 - 2x + 2.$ The tangent $r$, which is not horizontal, goes through the point $(2,1).$ What is the slope of $r$?

Question

Given a parabola $y= x^2 - 2x + 2$. The tangent r, which is not horizontal, goes trough the point (2,1). What is the slope of r?

I've already used the formula $\lim_{x\to 2}$$\frac{f(x)-f(2)}{x-2}$.

And then my result was that the slope is $2$ whereas the answer is $4$.

Can someone say what I've done wrong?

Answer 1

$(2,1)$ does not lie on the parabola

$$\dfrac{dy}{dx}=2x-2$$

So, the equation of the tangent at $(x_1,y_1=x_1^2-2x_1+2)$

$$\dfrac{y-(x_1^2-2x_1+2)}{x-x_1}=2x_1-2\ \ \ \ (1)$$

Now this passes through $(2,1)$

Use this in $(1)$ and solve for $x_1$

Answer 2

Hint...a line with gradient $m$ passing through $(2,1)$ is $$y-1=m(x-2)$$

If you solve this simultaneously with the given curve the resulting quadratic must have double roots, and this will give you the values of $m$