# Finding Coordinates for a point on a tangent line

I am given a equation $x^2-4x+3$ and told to identify the point $(a,f(a))$ at which the function has a tangent line with slope zero. How would I go about solving this? I don not know how to use derivatives.

• Without calculus, you can use your knowledge about parabolas. The parabola is $y=(x-2)^2-1$. Draw a picture. Our parabola reaches a minimum at $x=2$, and there the tangent line is flat. Oct 7, 2014 at 5:13

## 1 Answer

1 You are supposed to (know) how to derive to be given this question, If so :

$f(x) = x^2 -4x +3$

$f^(x) = 2x -4$

solving for the point with tangent 0 is same as solving for $f^(x)=0$

$f^`(x) = 0$

$2x - 4 = 0$

$2x = 4$

$x = 2$

2 Let's admit you don't (know) derivatives:

the slope of curve at any point can be calculated using the formula : $$Slope(x) = \frac{f(x+h) - f(x)}{h}$$ actually the above formula is valid when $h=0$ or in other words the slope is equal to the formula written above when h tends to 0

let's do that :

$\lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}$

$\lim_{h \rightarrow 0} \frac{(x+h)^2 -4(x+h) +3 -(x^2 - 4x +3)}{h}$

$\lim_{h \rightarrow 0} \frac{x^2 + 2xh +h^2 -4x -4h +3 - x^2 +4x -3}{h}$

$\lim_{h \rightarrow 0} \frac{2xh + h^2 -4h}{h}$

$\lim_{h \rightarrow 0} 2x + h -4$

$2x - 4$

Voila! it is the same result obtained by derivative method, no you solve for the slope equal to 0