Show that the tangent line to the curve y=x^3 at the point x=a also hits the curve at the point x=-2a.
Any help?! PLEASE!
dy/dx = 3x^2
at x=a , y = x^3 and dy/dx = 3a^2
equation of tangent:
y = 3a^2x + b
but (a,a^3) is on it, so
a^3 = 3a^2(a) + b
b = -2a^3
tangent equation: y = (3a^2)x - 2a^3
intersect that line with y = x^3
x^3 = 3a^2x - 2a^3
x^3 - 3a^2x + 2a^3 = 0
We already know that x-a is a factor, since x=a is a solution.
so it factors to
(x-a)(x^2 + ax - 2a^2) = 0
(x-a)(x-a)(x - 2a) = 0
so x=a, or x=a, or x = 2a
second line should have been ----
at x=a , y = x^3 and dy/dx = 3a^2
thank you soooooo much!
Sure! I'd be happy to help. To show that the tangent line to the curve y = x^3 at the point x = a also hits the curve at the point x = -2a, we need to find the equation of the tangent line and check if it intersects the curve at x = -2a.
To find the equation of the tangent line, we first need to find the derivative of the curve y = x^3 with respect to x. The derivative gives us the slope of the tangent line at any given point on the curve.
Using the power rule for differentiation, where the derivative of x^n is n*x^(n-1), we find that the derivative of y = x^3 is dy/dx = 3x^2.
Now that we have the slope of the tangent line, we can use the point-slope form of a linear equation to find the equation of the tangent line. The point-slope form is given by y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.
Since the point (a, a^3) lies on the curve y = x^3, we can use this point to find the equation of the tangent line. Plugging in the values, we have y - a^3 = 3a^2(x - a).
Now, to find out if the tangent line intersects the curve at x = -2a, we substitute x = -2a into the equation of the tangent line and check if the resulting y-value satisfies the equation of the curve.
Replacing x with -2a, we have y - a^3 = 3a^2(-2a - a), which simplifies to y - a^3 = 3a^2(-3a). Further simplifying, we get y - a^3 = -9a^3.
Combining like terms, we have y - a^3 + 9a^3 = 0, which simplifies to y + 8a^3 = 0.
Now we substitute x = -2a back into the equation of the curve y = x^3 to find the y-value at x = -2a. We get y = (-2a)^3 = -8a^3.
We can see that the y-values obtained from the tangent line equation and the curve equation are equal: y + 8a^3 = 0 and y = -8a^3.
Hence, the tangent line to the curve y = x^3 at the point x = a also intersects the curve at the point x = -2a.
Please let me know if there's anything else I can help you with!