Question

If $f\left(x\right)=x^2-4x+3$, then find the value of $f\left(2\right)-f(-1)+f\left(\frac{1}{2}\right)$.

Answer 1

$f\left(2\right)=2^2-4\left(2\right)+3=4-8+3=-1$

$f\left(1\right)=(-1{)}^2-4(-1)+3=1+4+3=8$

$f\left(\frac{1}{2}\right)={\left(\frac{1}{2}\right)}^2-4\left(\frac{1}{2}\right)+3=\frac{1}{4}-2+3=\frac{5}{4}$

$\Rightarrow f\left(2\right)-f\left(1\right)+f\left(\frac{1}{2}\right)=-1-8+\frac{5}{4}=-9+\frac{5}{4}=\frac{-31}{4}$.