Question

Write the equation on the tangent to the curve y = x² − x + 2 at the point where it crosses the y-axis.

Answer 1

When the curve crosses the y-axis, the point on the curve is of the form (0, y).
Here,
$$\begin{gathered} y=x^2-x+2 \\ \Rightarrow y=0-0+2=2 \\ \text{So, the point where the curve crosses the}\mathit{\text{y}}\text{-axis is (0, 2).} \\ \mathrm{Now}, \\ y=x^2-x+2 \\ \Rightarrow \frac{dy}{dx}=2x-1 \\ \text{Slope of the tangent,}\mathit{\text{m}}\text{=}{\left(\frac{dy}{dx}\right)}_{\left(0,2\right)}\text{=2}\left(0\right)\text{-1=-1} \\ \therefore \left(x_1,y_1\right)=\left(0,2\right) \\ \mathrm{and} \\ \text{Equation of tangent} \\ =y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-2=-1\left(x-0\right) \\ \Rightarrow y-2=-x \\ \Rightarrow x+y-2=0 \end{gathered}$$