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
Similar Questions
Find the equation of the normal to the curve y = \(2x^2 + 3 sin x\) at x = 0.
Find the angle between the curves xy = 6 and \(x^2 y\) =12.
Check the orthogonality of the curves \(y^2\) = x and \(x^2\) = y.
The angle of intersection between the curve \(x^2\) = 32y and \(y^2\) = 4x at point (16, 8) is
