# 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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