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14.02.2020 •
Mathematics
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=
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Ответ:
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
Ответ: