Question

Onde fica o vértice da parábola descrita pela função y=2x2+10x+8?

Answer 1

$\Large\boxed{\begin{array}{l}\sf y=2x^2+10x+8\\\sf \Delta=b^2-4ac\\\sf\Delta=10^2-4\cdot2\cdot8\\\sf\Delta=100-64\\\sf\Delta=36\\\sf x_V=-\dfrac{b}{2a}\\\\\sf x_V=-\dfrac{10}{2\cdot2}=-\dfrac{10\div2}{4\div2}=-\dfrac{5}{2}\\\\\sf y_V=-\dfrac{36}{4\cdot2}=-\dfrac{36\div4}{8\div4}=-\dfrac{9}{2}\\\\\sf\blue{ V\bigg(-\dfrac{5}{2},-\dfrac{9}{2}\bigg)}\end{array}}$