Find the equation of the normal to the curve y = 7x – 5x\(^2\) at x = 2
Explanation
y = 7x - 5x\(^2\)
\(\frac{dy}{dx}\) = slope/gradient
\(\frac{dy}{dx}\) = 7 - 10x at x = 2
m\(_1\) = - 13
But,the equation of normal, we need m\(_2\), from m\(_1\) m\(_2\) = - 1
m\(_2\) = \(\frac{-1}{m_1}\) = \(\frac{-1}{-13}\) = \(\frac{1}{13}\)
Using, \(\frac{ y - y_1}{x - x_1}\) = \(\frac{1}{m_2}\)
\(\frac{ y - y_1}{x - x_1}\) = \(\frac{1}{m_2}\)
To find the value of y we put x = 2 into y = 7x - 5x\(^2\) = 14 - 20 = -6
\(\frac{ y - (-6)}{x - 2}\) = \(\frac{1}{m_2}\)
y + 6 = \(\frac{1}{m_2}\)(x - 2)
13y + 78 = x - 2
13y - x + 80 = 0
Thus, the equation of normal = 13y - x + 80 = 0