[tex]\[ \begin{array}{l} \\ \rm y=x^{3}-3 x^{2}-4 x \\ \\ \displaystyle\rm\frac{d y}{d x}=3 x^{2}-6 x-4 \end{array} \][/tex]
[tex] \text{Slope of tangent }[/tex]
[tex] \rm\dfrac{d y}{d x}=3 x^{2}-6 x-4 \qquad\ldots \ldots(1)[/tex]
Given that tangent is parallel to 4x+y-3=0
[tex]\[ \begin{array}{l} \rm 4 x+y-3=0 \\ \\ \rm y=-4 x+3 \end{array} \][/tex]
Slope, m=-4 --------(2)
[tex] \tt \[ [\therefore y=m x+ C ] \] [/tex]
From (1) and (2),
[tex] \rm\[ 3 x^{2}-6 x-4=-4 \][/tex]
(Slope of parallel lines are equal)
[tex]\[ \begin{array}{l} \rm 3 x^{2}-6 x=0 \\ \\ \rm \quad 3 x(x-2)=0 \\ \\ \rm\quad x=0,2 \\ \\ \rm\quad y=x^{3}-3 x^{2}-4 x \\ \\ \rm \text { when } x=0, y=0 \\ \\ \rm \text { when } x=2, y=2^{3}-3(2)^{2}-4(2)=-12 \\ \\ \rm \therefore \quad \text { Required points are }(0,0) \text { and }(2,-12) \end{array} \][/tex]
[tex] \tiny\colorbox{lightyellow} {\text{ \bf♕ Brainliest answer }}[/tex][tex] \rule{300pt}{0.1pt}[/tex][tex]\mathbb\red{ \tiny A \scriptsize \: N \small \:S \large \: W \Large \:E \huge \: R}[/tex][tex] \rule{300pt}{0.1pt}[/tex]
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Answer :-
[tex]\[ \begin{array}{l} \\ \rm y=x^{3}-3 x^{2}-4 x \\ \\ \displaystyle\rm\frac{d y}{d x}=3 x^{2}-6 x-4 \end{array} \][/tex]
[tex] \text{Slope of tangent }[/tex]
[tex] \rm\dfrac{d y}{d x}=3 x^{2}-6 x-4 \qquad\ldots \ldots(1)[/tex]
Given that tangent is parallel to 4x+y-3=0
[tex]\[ \begin{array}{l} \rm 4 x+y-3=0 \\ \\ \rm y=-4 x+3 \end{array} \][/tex]
Slope, m=-4 --------(2)
[tex] \tt \[ [\therefore y=m x+ C ] \] [/tex]
From (1) and (2),
[tex] \rm\[ 3 x^{2}-6 x-4=-4 \][/tex]
(Slope of parallel lines are equal)
[tex]\[ \begin{array}{l} \rm 3 x^{2}-6 x=0 \\ \\ \rm \quad 3 x(x-2)=0 \\ \\ \rm\quad x=0,2 \\ \\ \rm\quad y=x^{3}-3 x^{2}-4 x \\ \\ \rm \text { when } x=0, y=0 \\ \\ \rm \text { when } x=2, y=2^{3}-3(2)^{2}-4(2)=-12 \\ \\ \rm \therefore \quad \text { Required points are }(0,0) \text { and }(2,-12) \end{array} \][/tex]
[tex] \tiny\colorbox{lightyellow} {\text{ \bf♕ Brainliest answer }}[/tex][tex] \rule{300pt}{0.1pt}[/tex][tex]\mathbb\red{ \tiny A \scriptsize \: N \small \:S \large \: W \Large \:E \huge \: R}[/tex][tex] \rule{300pt}{0.1pt}[/tex]
[tex]\[ \begin{array}{l} \\ \rm y=x^{3}-3 x^{2}-4 x \\ \\ \displaystyle\rm\frac{d y}{d x}=3 x^{2}-6 x-4 \end{array} \][/tex]
[tex] \text{Slope of tangent }[/tex]
[tex] \rm\dfrac{d y}{d x}=3 x^{2}-6 x-4 \qquad\ldots \ldots(1)[/tex]
Given that tangent is parallel to 4x+y-3=0
[tex]\[ \begin{array}{l} \rm 4 x+y-3=0 \\ \\ \rm y=-4 x+3 \end{array} \][/tex]
Slope, m=-4 --------(2)
[tex] \tt \[ [\therefore y=m x+ C ] \] [/tex]
From (1) and (2),
[tex] \rm\[ 3 x^{2}-6 x-4=-4 \][/tex]
(Slope of parallel lines are equal)
[tex]\[ \begin{array}{l} \rm 3 x^{2}-6 x=0 \\ \\ \rm \quad 3 x(x-2)=0 \\ \\ \rm\quad x=0,2 \\ \\ \rm\quad y=x^{3}-3 x^{2}-4 x \\ \\ \rm \text { when } x=0, y=0 \\ \\ \rm \text { when } x=2, y=2^{3}-3(2)^{2}-4(2)=-12 \\ \\ \rm \therefore \quad \text { Required points are }(0,0) \text { and }(2,-12) \end{array} \][/tex]