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Miscellaneous · Q32

Q.Find the equation of the tangent to the curve y=x2y=x^2 at the point (2,4)(2,4).

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✓ Free question

For y=x2y=x^2, the derivative (power rule, Section 10) is y′=2xy'=2x, so the slope of the tangent at x=2x=2 is y′(2)=2(2)=4y'(2)=2(2)=4 (Section 7). The point on the curve is (2,4)(2,4). Using the point-slope form y−y1=m(x−x1)y-y_1=m(x-x_1):

y−4=4(x−2)⟹y=4x−8+4=4x−4.y-4 = 4(x-2) \quad\Longrightarrow\quad y = 4x-8+4 = 4x-4.

✓Final answer

y=4x−4y = 4x-4

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