Question:

Find the equation of the tangent to $\displaystyle y = 8x^4 + 40x^3 + 26x^2 - 63x + 24$ at the point (3,-15). Find another point on the curve which has exactly the same equation for the tangent at the point.

Answer:

$\displaystyle f'(x) = 32x^3 + 120x^2 + 52x - 63$

$\displaystyle f'(-3) = m$

$\displaystyle m = -3$

$\displaystyle y - y_1 = m(x - x_1)$

$\displaystyle y = -3x + 6$

Now that I have the equation of the tangent, how do I go about finding the point? Do I have to set the equation of the tangent equal to the original equation and solve for $\displaystyle x$, or is there a simpler / more correct way of doing such?