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.


Can anyone tell me if I'm on the right path? I set the function and the slope of a line secant to the functions through the points (a, f(a)) and (-2a, f(-2a)) equal to each other and solved for a. I don't know what the a stands for. I got that a equals 1. I was thinking that a is the common slope of the tangent line and the function. Am I correct?

The a stands for any value. It could be a=2 and the tangent line at x=2 will hit the curve at x = -4. It could be a=3 and the tangent line at a=3 will hit the curve at -6, etc.
If it makes it easier, use some actual values. Such as, prove the tangent line at x=2 hits the curve at x=-4

Find the derivative: \frac{d}{dx}[x^{3}]=3x^{2}

Using the given coordinates, (a, a^{3}), find the equation of the tangent line:

a^{3}=3a^{2}(a)+b=3a^{3}+b

b=-2a^{3}

The tangent line equation is :

y=3a^{2}x-2a^{3}

Now, if x=-2a \Rightarrow{y=3a^{2}(-2a)-2a^{3}}=-8a^{3}

Try this - if you connect a line between the points (-2a, f(-2a)) and (a, f(a)), you can determine the slope of that line. Now, the tangent at f(a) is equal to its derivative there, so how does the slope of the line compare to f'(a)?

Also, I'm not sure what you meant by the word "secant" in the phrase "the slope of a line secant to the functions through the points (a, f(a)) and (-2a, f(-2a))" - please clarify. Thanks.

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.


Can anyone tell me if i'm on the right path? I set the function and the slope of a line secant to the functions through the points (a, f(a)) and (-2a, f(-2a)) equal to each other and solved for a. I don't know what the a stands for. I got that a equals 1. I was thinking that a is the common slope of the tangent line and the function. Am i correct?
Test 3 Solutions (http://www.math.temple.edu/~ludwick/teach/spring99/test3sol.html)

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