Question

Find the equation of tangent and normal to the curve $y=3x^3-4x+7$ at the point whose abscissa is 1.

Answer 1

$y=3x^3-4x+7$

$y^1=9x^2-4$

at $x=1$

$y^1=5$ $\to$ Slope of tangent

at $x=1$, $y3\left(1\right)-4+7$

$y=6$

$(x_1,y_1)=(1,6)$

Equation of tangent

$(y-6)=5(x-1)$

$5x-y+1=0$

Slope of normal$\left(m_1\right)$

$m_1{m}_2=-1$ $m_2=-\frac{1}{5}$

$(x_1,y_1)=(1,6)$Equation of normal

$(y-6)=-\frac{1}{5}(x-1)$

$x+5y-31=0$.