Question

F (x)=x^2-14+44 find the x value of the turning point

Answer 1

so... assuming
`f(x)=x^2-14x+44` then


`f(x)=x^2-14x+44\implies \begin{array}{lccclll} \boxed{1}x^2&\boxed{-14}x&\boxed{+44}\\ \uparrow &\uparrow &\uparrow\\ a&b&c \end{array} \\ \quad \\ \textit{vertex of a parabola}\\ \quad \\ y = {{ \boxed{a}}}x^2{{ \boxed{+b}}}x{{ \boxed{+c}}}\\ \quad \\ \left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)`