Question

If $f\left(x\right)$ is a quadratic polynomial such that graph of $y=f\left(x\right)$ touches at $(4,0)$ and intersects the positive $y-$axis at $4$, then which of the following is/are correct?

• A. $f\left(2\right)=1$
• B. $f\left(3\right)=\frac{1}{4}$
• C. $f\left(x\right)=\frac{1}{4}{x}^2-2x+4$
• D. $f\left(x\right)=\frac{1}{2}{x}^2-x+\frac{5}{2}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $f\left(2\right)=1$
B $f\left(3\right)=\frac{1}{4}$
C $f\left(x\right)=\frac{1}{4}{x}^2-2x+4$

Given $f\left(x\right)$ is a quadratic polynomial that touches the $x-$ axis at $x=4.$
$\therefore f\left(x\right)=0$ has equal roots equal to $x=4$, so
$f\left(x\right)=a(x-4{)}^2$
Now, the graph of $y=f\left(x\right)$ intersect on the positive $y-$axis
$f\left(0\right)=4\Rightarrow a(-4{)}^2=4\Rightarrow a=\frac{1}{4}$

Therefore , the required polynomial is,
$f\left(x\right)=\frac{1}{4}(x-4{)}^2$