Question

the equation of the tangent to the curve y=2x³-x²+2 at (1/2,2) is?​

Answer 1

Step-by-step explanation:

the equation of the curve is y=2x3-. ...

m1=m2.thus, the tangents to the given at the point where x=2 and x=-2 are parallel.

Answer 2

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☞Given that : y = 2x³ - x² + 2

Differentiate w.r.t. X

${\frac{dy}{dx} = \frac{d}{dx}(2</u><u>{</u><u>x}</u><u>^</u><u>{</u><u>3</u><u>}</u><u>- {x}^{2} + 2) = 6 {x}^{2} }$

${slope \: of \: tangent \: at \: ( \frac{1}{2}{ \: 2)}} = {m } \\ { m}= 6 ( { \frac{1}{2} )}^{2} { - 2 (\frac{1}{2})} \\ { therefore \: \: \: m = \frac{1}{2} } \\ {slope \: of \: normal \: at( \frac{1}{2} \: 2) } { = {m}^{</u><u>i</u><u>} = - 2}$

Equation of tangent is given by :

y - 2 = 1/2 ( x - 1/2 ) ➠ 2y - 4 = (2x-1)/ 2

4y - 8 = 2x - 1 ➠ 2x - 4y + 7 = 0

Equation of normal is given by :

y - 2 = -2( X-1/2 ) ➠ y - 2 = -2x + 1

2x + y - 3 = 0

If you find any problem in solution

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