find the point on the curve y=x^3+7x at which the normal line has the equation y=x+ 20
Answers
Given : y=x^3+7x at which the normal line has the equation y=x+ 20
To Find : The point :
Solution:
y = x³ + 7x
dy/dx = 3x² + 7
Slope of tangent = 3x² + 7
Slope of Normal = -1/(3x² + 7)
normal line has the equation y=x+ 20
=> slope = 1
Equate slope
1 = -1/(3x² + 7)
=> 3x² + 7 = - 1
=> 3x² = -8
=> x² = -8/3
Not possible
y = x³ + 7x
y=x+ 20
=> x³ + 7x = x + 20
=> x³ + 6x - 20 = 0
=> (x - 2) ( x² + 2x + 10) = 0
=> (x - 2) ( x² + 2x + 10) = 0
x = 2
y = 22
( 2, 22) is the point but its not normal to the equation
Looks some mistake in data
learn More:
If the slope of line joining the points (2, 5) and (x, 3) is 2, then find the ...
brainly.in/question/18032408
What is the slope of the line joining the points(2,0) and(-2,0) - Brainly ...
brainly.in/question/13724435