Question

How do you solve this to find the vertex f(x) = x^2-5x+4

Answer 1

`\bf \textit{vertex of a parabola}\\ \quad \\ \begin{array}{lcclll} f(x)=&1x^2&-5x&+4\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array}\qquad \left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)`