Question

What is the vertex of the graph of the function below? y=x^2-10x+24

Answer 1

`\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ \begin{array}{llcccl} y = & 1x^2& -10x& +24\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array} \qquad \left(-\cfrac{ b}{2 a}\quad ,\quad c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left( -\cfrac{-10}{2(1)}~~,~~24-\cfrac{(-10)^2}{4(1)} \right)\implies \left(5~~,~~24-25 \right)\implies (5~,~-1)`

Answer 2

The vertex of the equation is (5,-1). Rewrite in standard form. (h,k)