I am having trouble figuring out methodology to find another point of intersection for my tangent line $y=-38x+6$ and curve $2x^3+10x^2-28x$.
The question asks to determine tangent line at $x=-3$ for above function, and I know that they already intersect at $(-3, 120)$ having solved for $f(a)$. and using the secant method to find the slope of the above tangent line at $x=-3$.
Find another point of intersection:
$-38x+6 = 2x^3+10x^2-28x$
$0=2x^3+10x^2+10x-6$ [solve for roots]
I know that $(x+3)$ is a root by method of substitution for figuring $zero$ of the $cubic$, and when I perform the long div I end up with a remainder. It makes me think my methodology is incorrect.
How should I solve this one? Thanks
Best Answer
Note that $$2x^3+10x^2+10x-6=2 (x+3) \left(x^2+2 x-1\right)$$