Mathematics - Quadratic Equation Question with Solution | TestHub

The given expression is a quadratic $f(x)=x^2-4x+7$. Since the coefficient of $x^2$ is $1$ (which is positive), the parabola opens upwards, and thus has a minimum value. The minimum value occurs at the vertex, whose $x$ -coordinate is given by $x=-\frac{b}{2a}$. For this expression, $x=-\frac{-4}{2(1)}=2$. Substituting $x=2$ into the expression, we get the minimum value: $f(2)=(2{)}^2-4(2)+7=4-8+7=3$. Alternatively, by completing the square, $x^2-4x+7=(x-2{)}^2+3$, and since $(x-2{)}^2\ge 0$, the minimum value is $3$.