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

	Answered by bemath last updated on 28/Oct/20
	$⇒gradient = (dy/dx) ⇒3=4x+5 ; { ((x=−(1/2))),((y=(1/2)−(5/2)+1=−1)) :} Thus equation of tangent to curve y=2x^2 +5x+1 is 3x−y=−(3/2)+1 3x−y=−(1/2) or 6x−2y+1=0$
	$\Rightarrow g r a d i e n t = \frac{d y}{d x}$ $\Rightarrow 3 = 4 x + 5; {\begin{cases} x = - \frac{1}{2} \\ y = \frac{1}{2} - \frac{5}{2} + 1 = - 1 \end{cases}$ $T h u s e q u a t i o n o f t a n g e n t t o c u r v e$ $y = 2 x^{2} + 5 x + 1 i s 3 x - y = - \frac{3}{2} + 1$ $3 x - y = - \frac{1}{2} o r 6 x - 2 y + 1 = 0$
	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

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