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+1 vote

Find the equation of the tangent to the curve of y=2x^2+3x+1 at the point where x=1?

asked Jan 19, 2013 in CALCULUS by anonymous Apprentice

1 Answer

+2 votes

Equation of curve,y=2x^2+3x+1

At x=1




Slope of the tangent,m=dy/dx

Differentiate the equation of curve with respect to x


At x=1

Slope is m=4(1)+3


Equation of the tangent,


Substitute m=7,(x1,y1)=(1,6) in above equation


Simpplify : y-6=7x-7


Equation of the tangent is 7x-y=1


answered Jan 19, 2013 by bradely Mentor

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