Given: #f(x) = (x + 4)^2 - e^x " at "x = -3#
First find the slope of the curve at #x = -3#.
The first derivative is called the slope function. We need to find the first derivative:
#f'(x) = 2(x+4)^1 (1) - e^x#
#f'(x) = 2x + 8 - e^x#
slope = #m = f'(-3) = 2(-3) + 8 - e^-3 = 2 - 1/e^3 = (2e^3 - 1)/e^3#
A line that is normal is a perpendicular line. #m_("perpendicular") = -1/m: " "-e^3/(2e^3 - 1)#
#f(-3) = (-3 + 4)^2 - e^-3 = 1 - 1/e^3 = (e^3 - 1)/e^3#
To find the Normal line, use the point #(-3, (e^3 - 1)/e^3)#:
#y - y_1 = m (x - x_1)#
#y - (e^3 - 1)/e^3 = -e^3/(2e^3 - 1)(x +3)#
#y - (e^3 - 1)/e^3 = -e^3/(2e^3 - 1) x -(3e^3)/(2e^3 - 1)#
#y = -e^3/(2e^3 - 1) x -(3e^3)/(2e^3 - 1) +(e^3 - 1)/e^3#
#y = (e^3x)/(1-2e^3) +(3e^3)/(1-2e^3) +(e^3 - 1)/e^3#
Find a common denominator for the last two terms:
#y = (e^3x)/(1-2e^3) + (e^6 + 3e^3 - 1)/(e^3(1-2e^3)#
