Question-120036

Question Number 120036 by I want to learn more last updated on 28/Oct/20

Answered by bemath last updated on 28/Oct/20

\Rightarrow g r a d i e n t = \frac{d y}{d x}

\Rightarrow 3 = 4 x + 5; {\begin{cases} x = - \frac{1}{2} \\ y = \frac{1}{2} - \frac{5}{2} + 1 = - 1 \end{cases}

T h u s e q u a t i o n o f t a n g e n t t o c u r v e

y = 2 x^{2} + 5 x + 1 i s 3 x - y = - \frac{3}{2} + 1

3 x - y = - \frac{1}{2} o r 6 x - 2 y + 1 = 0

Commented by I want to learn more last updated on 29/Oct/20

Thanks sir

Commented by I want to learn more last updated on 29/Oct/20

Sir, how did you get 3 x - y from y = {2 x}^{2} + 5 x + 1