# Write the equation of the tangent to the curve

Question:

Write the equation of the tangent to the curve $y=x^{2}-x+2$ at the point where it crosses the $y$-axis.

Solution:

Given that the curve $y=x^{2}-x+2$ has a point crosses the $y$-axis.

The curve will be in the form of $(0, y)$

$\Rightarrow y=0-0+2$

$\Rightarrow y=2$

So, the point at which curve crosses the $y$-axis is $(0,2)$.

Now, differentiating the equation of curve w.r.t. $x$

$\frac{d y}{d x}=x-1$

For $(0,2)$,

$\frac{\mathrm{dy}}{\mathrm{dx}}=-1$

Equation of the tangent:

$\left(\mathrm{y}-\mathrm{y}_{1}\right)=$ Slope of tangent $\times\left(\mathrm{x}-\mathrm{x}_{1}\right)$

$\Rightarrow(y-2)=-1 \times(x-0)$

$\Rightarrow y-2=-x$

$\Rightarrow x+y=2$