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,

\[y = x^2 - x + 2\]

\[ \Rightarrow y = 0 - 0 + 2 = 2\]

\[\text { So, the point where the curve crosses the y-axis is }(0, 2).\]

\[\text { Now,} \]

\[y = x^2 - x + 2\]

\[ \Rightarrow \frac{dy}{dx} = 2x - 1\]

\[\text { Slope of the tangent },m= \left( \frac{dy}{dx} \right)_\left( 0, 2 \right) =2\left( 0 \right)-1=-1\]

\[ \therefore \left( x_1 , y_1 \right) = \left( 0, 2 \right)\]

\[\text { and }\]

\[\text { Equation of tangent }\]

\[ = y - y_1 = m\left( x - x_1 \right)\]

\[ \Rightarrow y - 2 = - 1\left( x - 0 \right)\]

\[ \Rightarrow y - 2 = - x\]

\[ \Rightarrow x + y - 2 = 0\]