At what point is $y=5x+965$ to the graph of $y=2x^3-3x^2-247x-7$?

Question

Here is the question:

The line $y=5x+965$ is tangent to the graph of $y=2x^3-3x^2-247x-7$ at the point...?

Anytime I am trying to plot the graph I end up with a line that's basically attached to the $y$-axis.

Can anyone help?

Answer 1

Hint:

Substituting the $y$ of the line in the cubic we have: $$ 2x^3-3x^2-252x-972=0 $$ that factorize as $(x+6)^2(2x-27)=0$

So the double root $x=-6 $ gives the point of tangency at $P=(-6,935)$. (Difficult to visualize in a graph)

Answer 2

You have the two lines $$y=5x+965$$ and $$y=2x^3-3x^2-247x-7$$ Since you are trying to find where they intersect, you can substitute the right hand side of the second equation for the $y$ in the first equation. You get $$5x+965=2x^3-3x^2-247x-7$$ which simplifies to $$2x^3-3x^2-252x-972$$ You can factor this into $$(x+6)^2(2x-27)=0$$

The roots are $x=-6$ and $x=13.5$, and substituting them into the first you get $(-6, 935)$ and $(13.5, 1032.5)$ as the points of intersection.

Answer 3

Alternative answer is that you can consider the line y=5x+965 on the form y=kx+m

where k is = 5. Then the derivative of $f(x)=2x^3-3x^2-247x-7$ needs to also be 5 in that point.

This means you can set $f'(x)=5 $ and get $6x^2-6x-252=0\Leftrightarrow x_1=-6$ $,x_2=7$

Then you can find f(-6) and f(7). The equation for a tangentline in a point is:

$y-y_1 = k(x-x_1)$ which gives $y=5(x-(-6))+f(-6)$ which is your tangentline.

$y=5(x-7)+f(7)$ is a parallell line to y = 5x+965. with the equation y = 5x -1232