Question:
Find the point on the curve $y=x^{2}-2 x+3$, where the tangent is parallel to $x$-axis.
Solution:
Given curve $y=x^{2}-2 x+3$
We know that the slope of the $x$-axis is 0 .
Let the required point be $(a, b)$.
$\because$ the point lies on the given curve
$\therefore b=a^{2}-2 a+3$ .....(1)
Now, $y=x^{2}-2 x+3$
$\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}-2$
Slope of the tangent at $(a, b)=2 a-2$
According to the question,
$2 a-2=0$
$\Rightarrow a=1$
Putting this in (1),
$b=1-2+3$
$\Rightarrow b=2$
So, the required point is $(1,2)$