Answer
There are two inflection points when
$c \lt -2\sqrt{\frac{2}{3}}~~$ or $~~c \gt 2\sqrt{\frac{2}{3}}$
There is one inflection point when
$c = -2\sqrt{\frac{2}{3}}~~$ or $~~c = 2\sqrt{\frac{2}{3}}$
There are no inflection points when
$-2\sqrt{\frac{2}{3}} \lt c \lt 2\sqrt{\frac{2}{3}}$
As $c$ decreases, $P(x)$ does not fall as far below the x-axis, and as $c$ continues to decrease, $P(x)$ does not fall below the x-axis at all.
Work Step by Step
$P(x) = x^4+cx^3+x^2$
$P'(x) = 4x^3+3cx^2+2x$
$P''(x) = 12x^2+6cx+2$
We can find the values of $x$ such that $P''(x) = 0$:
$P''(x) = 12x^2+6cx+2 = 0$
$6x^2+3cx+1 = 0$
We can use the quadratic formula:
$x = \frac{-3c\pm \sqrt{(3c)^2-(4)(6)(1)}}{2(6)}$
$x = \frac{-3c\pm \sqrt{9c^2-24}}{12}$
$x = -\frac{c}{4}\pm \frac{\sqrt{9c^2-24}}{12}$
There are two inflection points when $9c^2 \gt 24$:
$c^2 \gt \frac{8}{3}$
$c \lt -2\sqrt{\frac{2}{3}}~~$ or $~~c \gt 2\sqrt{\frac{2}{3}}$
There is one inflection point when $9c^2 = 24$:
$c^2 = \frac{8}{3}$
$c = -2\sqrt{\frac{2}{3}}~~$ or $~~c = 2\sqrt{\frac{2}{3}}$
There are no inflection points when $9c^2 \lt 24$:
$c^2 \lt \frac{8}{3}$
$-2\sqrt{\frac{2}{3}} \lt c \lt 2\sqrt{\frac{2}{3}}$
We can see a sketch of four graphs of $P(x)$ when $c = 1,2,3,4$
As $c$ decreases, $P(x)$ does not fall as far below the x-axis, and as $c$ continues to decrease, $P(x)$ does not fall below the x-axis at all.
