How do you find the equation of the tangent and normal line to the curve y=x2+2x+3 at x=1?

Question

Emily Green · Accepted Answer

y=x2+2x+3at x=1,y=12+2(1)=3=6let say m1=gradient of tangent.dydx=2x+2at x=1,m1=dydx=2(1)+2=4Therefore the equation of tangent at (1,6),(y−6)=m1(x−1)(y-6)=4(x-1)y=4x-4+6y=4x+2let say m2= gradient of normal.m1⋅m2=−1m2=−14Therefore the equation of normal at (1,6),(y−6)=m2(x−1)(y−6)=−14(x−1)y=−14x+14+6y=−14x+2544y=-x+25

fallendreamsit2p · Accepted Answer

f(x)=x2+2x+3 and x0=1Find the value of the function at the given point: y0=f(1)=6The slope of the normal line at x=x0 is the negative reciprocal of the derivative of the function, evaluated at x=x0:M(x0)=−1f′(x0)Find the derivative: f′(x)=(x2+2x+3)′=2(x+1)Hence, M(x0)=−1f′(x0)=−12(x0+1)Next, find the slope at the given point.m=M(1)=−14Finally, the equation of the normal line is y−y0=m(x−x0)Plugging the found values, we get that y−6=−x−14Or, more simply: y=254−x4.

How do you find the equation of the

Answered question

Answer & Explanation

New Questions in Calculus 1