Find equation of a tangent

Question

Find equation of a tangent which was drawn from $(1,-2)$ for $f(x)=x^2 - 2x$ [ This is my try but I can't continue ,maybe I have to find $(A,B)$ first ] Thanks

Answer 1

Without making reference to the Taylor series (like the current accepted answer), you can approach it like this.

$f'(x) = 2(x-1)$

So the slope of the line will be given by $2(A-1)$

The equation of a line that passes through $(A,B)$ with slope $2(A-1)$ is given by:

$y -B = 2(A-1)(x-A)$

Noting that the line also passes through $(1,-2)$, we get:

$-2-B = 2(A-1)(1-A)$ $B = 2(A-1)^2-2 = 2((A-1)^2 - 1) = 2(A)(A-2)$

Now consider that the point $(A,B)$ also lies on the curve, so we get:

$B = A(A-2)$.

Equating the two, we get $B=2B$, which gives $B=0$, so $A(A-2) = 0$

Hence $A=0$ or $A=2$.

All that's left is to find the possible tangent lines.

One passes through $(0,0)$ and $(1,-2)$, so its equation is:

$y = -2x$

The other passes through $(2,0)$ and $(1,-2)$ so its equation is:

$y = 2(x-2)$

$y = 2x-4$

Answer 2

A line that goes through $(1,-2)$ has equation $$y=a(x-1)-2$$

Now we need the equation $$x^2-2x=a(x-1)-2$$

to have one unique solution, namely the point where the line is tangent to the parabola.

Use the discriminant of this quadratic equation to find $a$ and therefore the equation of the tangent line.

Answer 3

The 1^st order taylor series expansion of $f(x) = x^2 - 2 x $ about some arbitrary point $x=a$ is:

$$ y = {\rm taylor}(f(x),x=a,1) = 2 x (a-1)-a^2 $$

Now this line must pass through $(1,-2)$ or $$ -2 = 2 (1) (a-1)-a^2 \Rightarrow \begin{cases} a=0 \\ a=2 \end{cases} $$

So the answer is $y=-2 x$ or $y=2 x-4$ for the tangents.