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Question 131 of 160

Q.The equation of tangent to the curve y=x2+4x+1y = x^2 + 4x + 1 at (−1,−2)(-1, -2) is

(a) 2x−y=02x - y = 0
(b) 2x+y−5=02x + y - 5 = 0
(c) 2x−y−1=02x - y - 1 = 0
(d) x+y−1=0x + y - 1 = 0
Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2016MCQ· 2mImportance★★★★★
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Compute dydx\dfrac{dy}{dx} at x=−1x=-1 for the slope, then use the point-slope form.

y=x2+4x+1⇒dydx=2x+4y=x^2+4x+1 \Rightarrow \frac{dy}{dx}=2x+4

At (−1,−2)(-1,-2): slope =2(−1)+4=2=2(-1)+4=2.

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