Question

If $f\left(x\right)$ is a polynamial function of the second degree such that $f(-3)=6,f\left(0\right)=6$ and $f\left(2\right)=11$, then the graph of the function $f\left(x\right)$ cuts the ordinate $x=1$ at the point

• A. $(1,8)$
• B. $(1,4)$
• C. $(1,-2)$
• D. none of these

Answer 1

The correct option is A $(1,8)$

Let $f\left(x\right)=a{x}^2+bx+c$
Now, $f\left(0\right)=6$
$\Rightarrow c=6$.
$\Rightarrow f\left(x\right)=a{x}^2+bx+6$
Given : $f(-3)=6$
$\Rightarrow 9a-3b+6=6$
$b=3a$ ...(i)
Hence, $f\left(x\right)$ now becomes
$f\left(x\right)=a{x}^2+3ax+6$
Also, $f\left(2\right)=11$
$\Rightarrow 4a+6a+6=11$
$\Rightarrow 10a=5$
$\Rightarrow a=\frac{1}{2}$
Thus, we have
$f\left(x\right)=\frac{x^2}{2}+\frac{3x}{2}+6$
$f\left(1\right)$ $=\frac{1}{2}+\frac{3}{2}+6$
$=6+2$
$=8$
Hence, the ordered pair will be $(1,8)$.