the tangent line to the curve y=1/3(x^3)-3x^2+11x+15 is parallel to the line 9x-3y=2 at two points on the curve.Find the two points.
To find the two points where the tangent line to the curve is parallel to the line 9x - 3y = 2, we need to follow these steps:
Step 1: Find the slope of the given line.
The equation of the given line is in the form Ax + By = C. To find the slope, we need to rearrange the equation in the form y = mx + b, where m is the slope. In this case, we have:
9x - 3y = 2
-3y = -9x + 2
Divide both sides by -3:
y = (9/3)x - 2/3
y = 3x - 2/3
Therefore, the slope of the given line is 3.
Step 2: Find the derivative of the given curve.
The equation of the curve is y = (1/3)x^3 - 3x^2 + 11x + 15. To find the derivative, we need to differentiate the equation with respect to x. Differentiating each term separately, we get:
dy/dx = d/dx[(1/3)x^3] - d/dx[3x^2] + d/dx[11x] + d/dx[15]
dy/dx = (1/3)(3x^2) - 2(3x) + 11 + 0
dy/dx = x^2 - 6x + 11
Step 3: Find the x-values where the tangent line is parallel to the given line.
To find the x-values, we need to set the derivative equal to the slope of the given line (since parallel lines have equal slopes):
x^2 - 6x + 11 = 3
x^2 - 6x + 11 - 3 = 0
x^2 - 6x + 8 = 0
Step 4: Solve the quadratic equation for x.
We can solve the quadratic equation by factoring or using the quadratic formula. In this case, the equation cannot be factored easily, so we'll use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
Using the quadratic formula, with a = 1, b = -6, and c = 8, we can plug in these values and solve for x:
x = (-(-6) ± √((-6)^2 - 4(1)(8))) / 2(1)
x = (6 ± √(36 - 32)) / 2
x = (6 ± √4) / 2
x = (6 ± 2) / 2
x = (6 + 2) / 2 or (6 - 2) / 2
x = 8 / 2 or 4 / 2
x = 4 or 2
Therefore, the two points where the tangent line is parallel to the given line on the curve y = (1/3)x^3 - 3x^2 + 11x + 15 are (4, f(4)) and (2, f(2)), where f(x) is the equation of the curve. To find the corresponding y-values, substitute the x-values into the equation of the curve and evaluate:
f(4) = (1/3)(4^3) - 3(4^2) + 11(4) + 15
f(4) = (1/3)(64) - 3(16) + 44 + 15
f(4) = 21.33 - 48 + 44 + 15
f(4) = 32.33
f(2) = (1/3)(2^3) - 3(2^2) + 11(2) + 15
f(2) = (1/3)(8) - 3(4) + 22 + 15
f(2) = 2.67 - 12 + 22 + 15
f(2) = 27.67
Therefore, the two points are (4, 32.33) and (2, 27.67).