At how many points on the curve

Question

y=4x^5-3x^4+15x^2+6 will the line tangent to the curve pass through the origin?

Answer 1

when the tangent passes through the origin, it is of the form

y = kx

So, we want places where the slope is y/x.

y' = 20x^4 - 12x^3 + 30x

So, we want places where

20x^4 - 12x^3 + 30x = (4x^5-3x^4+15x^2+6)/x
That is, where
20x^5-12x^4+30x^2 = 4x^5-3x^4+15x^2+6
16x^5-9x^4+15x^2-6 = 0
The only real root to that is x=0.62446

So, since y(.62446) = 11.7729

So, at (.62446,11.7729) the tangent line is

y = 18.85x

See the graphs at

http://www.wolframalpha.com/input/?i=plot+y+%3D+4x^5-3x^4%2B15x^2%2B6+and+y+%3D+18.85x+for+x+%3D+-1+to+1

If you look at the graph at

http://www.wolframalpha.com/input/?i=plot+y+%3D+4x^5-3x^4%2B15x^2%2B6+

you might wonder whether the tangent line near the point of inflection passes through (0,0). Let's see.

y' = 20x^4 - 12x^3 + 30x
y" = 80x^3 - 36x^2 + 30
y"=0 at (-.5981,10.6758)
y'(-.5981) = -12.8162
So, the tangent through that point is
y=-12.8162(x+.5981)+10.6758

This line misses (0,0), passing through (0.2349,0)

Since the slope is less on both sides of the inflection point, the tangents nearby cannot pass through (0,0).

you can see from the graphs at

http://www.wolframalpha.com/input/?i=plot+y%3D4x^5-3x^4%2B15x^2%2B6+and+y%3D-12.8162%28x%2B.5981%29%2B10.6758

that this is so. So, the first solution is the only one.