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

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Answer 1 · 2017-03-08T04:55:29

a. Equation of tangent, # y =4 x + 2#
b. Equation of normal, #4 y = -x +25#

#y = x^2 + 2 x + 3#

at #x =1#, #y = 1^2+ 2(1) + 3 =6#

let say #m_1# = gradient of tangent.
#(d y)/(d x) = 2 x + 2#

at #x =1#, #m_1 = (d y)/(d x) = 2 (1) + 2 =4#

Therefore the equation of tangent at (1,6),
#(y-6) = m_1(x-1)#
#(y-6) = 4(x-1)#
# y =4 x -4 +6#
# y =4 x + 2#

let say #m_2# = gradient of normal.
#m_1 * m_2 = -1#
#m_2 = -1/4#

Therefore the equation of normal at (1,6),
#(y-6) = m_2(x-1)#
#(y-6) = -1/4(x-1)#

#y = -1/4 x + 1/4 +6#

#y = -1/4 x + 25/4#

#4 y = -x +25#