Mathematics: What is the vertex of h=-16t^2+20t+6

What is the vertex of h=-16t^2+20t+6 - id1622063013275, scasemere

Ответ:

mauralawson

05.01.2022

$\large\boxed{\left(\dfrac{5}{8},\ 12\dfrac{1}{4}\right)}$

Step-by-step explanation:

$y=ax^2bx+c\\\\(h,\ k)-vertex\\\\h=\dfrac{-b}{2a}\\\\k=\dfrac{-(b^2-4ac)}{4a}\\---------------\\\text{We have}\\\\h=-16t^2+20t+6\\\\a=-16,\ b=20,\ c=6\\\\h=\dfrac{-20}{2(-16)}=\dfrac{-20}{-32}=\dfrac{20:4}{32:4}=\dfrac{5}{8}\\\\k=\dfrac{-(20^2-4(-16)(6))}{4(-16)}=\dfrac{-(400+384)}{-64}=\dfrac{784}{64}=12\dfrac{1}{4}\\\\\text{The vertex}\ \left(\dfrac{5}{8},\ 12\dfrac{1}{4}\right)$