Question

Find the equation of the tangent line and normal line to the curve y=(6+4x)2 at the point (6,900). Write the equations of the lines in the form y=mx+b. Tangent line: y= equation editor Equation Editor Normal line: y=

Answer 1

Answer:

Equation of the tangent to the curve

y = 240x - 215994

Equation of the normal

y = (-1/240)x + 9.75 = - 0.00417x + 9.75

Step-by-step explanation:

y = (6 + 4x)² = 36 + 48x + 16x² = 16x² + 48x + 36

dy/dx = 32x + 48

At the point (6,900),

dy/dx = 32(6) + 48 = 240

Equation of the tangent at point (a,b) is

(y - b) = m(x - a)

a = 6, b = 900, m = 240

y - 6 = 240(x - 900)

In the y = mx + b form,

y - 6 = 240x - 216000

y = 240x - 215994

The slope of the normal line = -(1/slope of the tangent line) (since they're both perpenducular to each other)

Slope of the normal line = -1/240

Equation of normal

y - 6 = (-1/240)(x - 900)

y - 6 = (-x/240) + 3.75

y = (-1/240)x + 9.75

y = - 0.00417x + 9.75