Question

What is the vertex of h=-16t^2+20t+6​

Answer 1

`\large\boxed{\left((5)/(8),\ 12(1)/(4)\right)}`

Explanation:

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