Question

Suppose $a,b,c$ are three distinct real numbers and $p\left(x\right)$ is a real quadratic polynomial such that
$\left[\begin{array}{ccc}4a^2& 4a& 1\\ 4b^2& 4b& 1\\ 4c^2& 4c& 1\end{array}\right]\left[\begin{array}{c}p(-1)\\ p\left(1\right)\\ p\left(1\right)\end{array}\right]=\left[\begin{array}{c}3a^2+3a\\ 3b^2+3b\\ 3c^2+3c\end{array}\right]$

Answer 1

$a,b,c$ are three roots of the quadratic equation
$\left(4f\right(-1)-3)x^2+\left(4f\right(1)-3))x+f(2)=0\Rightarrow 4f(-1)=3,4f(1)=3andf(2)=0$.
Let $f\left(x\right)=(x-2)(Ax+B)$
Now, $3=4f(-1)=4(-3)(-A+B)\Rightarrow A-B=1/4$
$3=4f(-1)=4(-1)(A+B)\Rightarrow A+B=-3/4$
$\therefore A=-1/4,B=-2/4$
Thus, $f\left(x\right)=\frac{1}{4}(4-x{)}^2$.
The graph of $y=f\left(x\right)$ is given in Fig.
A) $x$ coordinates of points of intersection of $y=f\left(x\right)$ when the $x$-axis are $\pm 2$.
B) $Area\frac{3}{2}=\stackrel{2}{\underset{-2}{\int }}\frac{1}{4}(4-x^2)dx=\frac{3}{8}\left(2\right){\left[\begin{array}{c}\begin{array}{c}4x-\frac{x^2}{3}\end{array}]\end{array}\right]}_0^2=4$
C) Maximum value of $f\left(x\right)$ is 1
D) Length of intercept on the $x$-axis is 4.