How do you find the equation of the tangent line and normal line to the curve #y = (5+6x)^2# at the point (-4,361)?

Question

Samantha Adkison · Accepted Answer

1) Equation of Tangent Step1: The tangent is a line so it has the form of a line:
# y - y_0 = m(x - x_0)#
#(x_0, y_0)# represents the point at which the tangent touches the curve, in this case we have #(-4, 361)#
Also the gradient, #m = (dy)/(dx)#
So,
# y = (5 + 6x)^2#
#=> (dy)/(dx) = 2*(5 + 6x)xx6#
#           = 12(5 + 6x)# Step2: At the point #(-4, 361)# we substitute the value of #x = -4# and this becomes, # (dy)/(dx) = 12(5 + 6*(-4))#
# = 12*(-19) = -228# So, # m = -228# Step3: Putting the point #(-4, 361)# and gradient # m = -228# in the slope-point form: # => y - 361 = -288(x - (-4))#
hence equation of tangent is # y - 1 = -288(x + 4)# Equation of normal The equation of normal is also a line of the same form as the tangent. The only difference is at the level of the gradient. We can deduce the gradient of the normal from the fact that the normal is a line perpendicular to the tangent This implies the product of their gradients is #-1# So let #m_N# be gradient of normal and #m_T# be gradient of tangent. # => m_N xx m_T = -1#
# => m_N = -1/(m_T)   = -1/(-288) = 1/288# Hence equation of normal is   #   y -361 = (1/288)xx(x + 4)#

Christopher Allers · Answer

undefined To find the equation of the tangent line and normal line to the curve y = (5+6x)^2 at the point (-4,361), we need to find the derivative of the function at that point. The derivative of y = (5+6x)^2 is dy/dx = 12(5+6x).

To find the slope of the tangent line, substitute x = -4 into the derivative: dy/dx = 12(5+6(-4)) = -72.

The slope of the tangent line is -72.

To find the equation of the tangent line, use the point-slope form: y - y1 = m(x - x1), where (x1, y1) is the given point (-4,361).

Substituting the values, we have y - 361 = -72(x - (-4)). Simplifying, we get y = -72x + 23.

The equation of the tangent line is y = -72x + 23.

To find the slope of the normal line, we take the negative reciprocal of the slope of the tangent line. The negative reciprocal of -72 is 1/72.

To find the equation of the normal line, we use the point-slope form again: y - y1 = m(x - x1). Substituting the values, we have y - 361 = (1/72)(x - (-4)). Simplifying, we get y = (1/72)x + 361.

The equation of the normal line is y = (1/72)x + 361.