The tangent to y=x³-6x²+ 8x at (3, -3) intersects the curve at another point. Determine this point.
To find the point of intersection, we need to find the equation of the tangent line and then solve it simultaneously with the equation of the curve.
First, let's find the equation of the tangent line using the derivative of the curve:
y = x³ - 6x² + 8x
Taking the derivative with respect to x:
dy/dx = 3x² - 12x + 8
To find the slope of the tangent line at the point (3, -3), substitute x = 3 into the derivative:
m = dy/dx = 3(3)² - 12(3) + 8 = 27 - 36 + 8 = -1
Hence, the slope of the tangent line is -1. Using the point-slope form of a linear equation, the equation of the tangent line is:
y - y₁ = m(x - x₁)
where (x₁, y₁) is the point (3, -3) and m is the slope:
y + 3 = -1(x - 3)
Simplifying, we get:
y + 3 = -x + 3
y = -x
Now we have the equation of the tangent line. To find the point of intersection, we substitute this equation into the equation of the curve:
y = x³ - 6x² + 8x
-x = x³ - 6x² + 8x
Rearranging the terms, we get:
x³ - 6x² + 9x = 0
Since x is a common factor, we can factor out an x:
x(x² - 6x + 9) = 0
x(x - 3)(x - 3) = 0
This gives us three possible solutions for x: x = 0, x = 3, x = 3. However, since x = 0 does not intersect the curve again, we can disregard it. Therefore, the point of intersection is when x = 3.
To find the y-coordinate, substitute x = 3 into the equation of the curve:
y = (3)³ - 6(3)² + 8(3)
y = 27 - 54 + 24
y = -3
So, the point of intersection is (3, -3).
Therefore, the tangent line to the curve y = x³ - 6x² + 8x at (3, -3) intersects the curve again at the point (3, -3).