Question

If $f\left(x\right)$ is a polynomial function of the second degree such that, $f(-3)=6,f\left(0\right)=6andf\left(2\right)=11,$ 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

$(1,8)$

Explanation for correct option:

Step 1: Find the value of $a,b$ and $c$.

Given the polynomial function of the second degree

$f\left(x\right)=a{x}^2+bx+c$

Substitute $x=0$, we get

$$\begin{gathered} f\left(0\right)=6 \\ \Rightarrow c=6 \end{gathered}$$

Now,

$$\begin{gathered} f\left(2\right)=11 \\ \Rightarrow 4a+2b+6=11 \\ \Rightarrow 4a+2b=5...\left(1\right) \end{gathered}$$

And

$$\begin{gathered} f(-3)=6 \\ \Rightarrow 9a–3b+6=6 \\ \Rightarrow 9a=3b \\ \Rightarrow b=3a...\left(2\right) \end{gathered}$$

Substitute $\left(2\right)$ in $\left(1\right)$, we get,

$$\begin{gathered} 4a+6a=5 \\ \Rightarrow 10a=5 \\ \Rightarrow a=\frac{1}{2} \\ \Rightarrow b=\frac{3}{2}\left[\mathrm{from}2\right] \\ c=6 \end{gathered}$$

Step 2: Find the coordinate:

Substituting these values in the quadratic equation we get,

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

At $x=1$,

$$\begin{gathered} f\left(1\right)=\frac{1}{2}+\frac{3}{2}+6 \\ =2+6 \\ =8 \end{gathered}$$

Hence, the required coordinate is $(1,8)$

Hence, the correct option is A.