SOLUTION:
The equation of a curve is given by
y = 6x^5+12
Obtain the tangent line to the curve at the point where x = 3. Please enter your answer as an equation in the form: y = m x +
Algebra ->
Quadratic Equations and Parabolas
-> SOLUTION:
The equation of a curve is given by
y = 6x^5+12
Obtain the tangent line to the curve at the point where x = 3. Please enter your answer as an equation in the form: y = m x +
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Question 869495:
The equation of a curve is given by
y = 6x^5+12
Obtain the tangent line to the curve at the point where x = 3. Please enter your answer as an equation in the form: y = m x + c
for some constants m, c. Answer by htmentor(1343) (Show Source):
You can put this solution on YOUR website! The slope of the tangent line at x=3 is given by the derivative, dy/dx:
dy/dx = 30x^4 = 30*3^4 = 2430
So the line has the form y = 2430x + c
The value of y at x=3 is given from the original curve:
y = 6*3^5 + 12 = 1470
Another point on the line is (0,c)
So m = (1470-c)/3 = 2430
c = -2430*3 + 1470 = -5820
Ans: y = 2430x - 5820
