Find the point on the curve

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 $(1,2)$

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