Let $f(x)=x^4+(2-a) x^3-(2 a+1) x^2+(a-2) x+2 a$ for some $a \geq 2$. Draw two tangent lines of its graph at the point $(-1,0)$ and $(1,0)$ and let $P$ be the intersection point. Denote by $T$ the area of the triangle whose vertices are $(-1,0),(1,0)$ and $P$. Let $A$ be the area of the domain enclosed by the interval $[-1,1]$ and the graph of the function on this interval. Show that $T \leq 3 A / 2$.
To solve this problem, I'll start by finding the equation of the tangent lines at the points $(-1,0)$ and $(1,0)$ on the graph of the function $f(x)$. Then, I'll find the coordinates of the intersection point $P$ and use them to calculate the area of the triangle formed by $(-1,0),(1,0)$, and $\mathrm{P}$, denoted as T. Finally, I'll calculate the area $A$ enclosed by the interval $[-1,1]$ and the graph of $f(x)$ and prove that $T \leq \frac{3}{2} A$.
Is this approach correct?
Hints (for finding $T$ and $A$):
Calculus is your friend. To find $A$, i.e. the area of the region which is bounded by the curve $y=f(x)$, the $x-$ axis, and the parallel lines $x=a$ and $x=b$, use definite integral: $$A=\int_a^bf(x) \, \mathrm dx.$$
To find equations of the tangents $t_1$ and $t_2$ to a curve $y=f(x)$ at a given points use well known formula: $$y-y_0=f'(x_0)(x-x_0).$$
Now, when you have both tangents you should be able to find its intersection $P=(x_3,y_3)$.
Finally, to find $T$, i.e. the area of triangle $\triangle P_1P_2P$ use yet another well known formula: $$T=\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|.$$