Question

If $f\left(x\right)$ is a polynomial 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 we have

$f(-3)=6$

$\Rightarrow a{(-3)}^2+(-3)b+c=6$

$9a-3b+c=6⟶\left(1\right)$

Also we have, $f\left(0\right)=6$

So, $a\left(0\right)+b\left(0\right)+c=6$

$\Rightarrow c=6⟶\left(2\right)$

Also, $f\left(2\right)=11$

$a{\left(2\right)}^2+b\left(2\right)+c=11$

$4a+2b+c=11⟶\left(3\right)$

So from$\left(1\right)\&\left(2\right)$ ,we get:

$\Rightarrow 9a-3b+6=6$

$\Rightarrow 9a=3b,orb=3a⟶\left(4\right)$

Putting equation (4) in (5) and equation (2) in (3), we get:

$\Rightarrow 4a+2\left(3a\right)+6=11$

$\Rightarrow 10a+6=11ora=\frac{5}{10}or\frac{1}{2}$

From (4), we have,

$b=3\times \frac{1}{2}=\frac{3}{2}$

So, $f\left(x\right)=\frac{1}{2}{x}^2+\frac{3}{2}x+6$

Now, $f(x=1)=\frac{1}{2}{\left(1\right)}^2+\frac{3}{2}\left(1\right)+6$

$\Rightarrow \frac{1}{2}+\frac{3}{2}+6$

$\Rightarrow$ 8

So point of intersection of $f\left(x\right)$ & $x=1$ is $(1,8)$