Find the equation of the tangent to curve y = \(-5x^2 + 6x + 7\)  at the point (1/2, 35/4).

Solution :

The equation of the given curve is

y = \(-5x^2 + 6x + 7\)

\(\implies\) \(dy\over dx\) = -10x + 6

\(\implies\) \(({dy\over dx})_{(1/2, 35/4)}\) = \(-10\over 4\) + 6 = 1

The required equation at (1/2, 35/4) is

y – \(35\over 4\) = \(({dy\over dx})_{(1/2, 35/4)}\) \((x – {1\over 2})\)

\(\implies\) y – 35/4 = 1(x – 1/2)

\(\implies\) Equation of tangent is y = x + 33/4


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