Question

Please help!!!
Find the vertex and zeros of the parabola

Answer 1

Vertex:
`((1)/(2) ,16)`

Zeros:
`(3)/(2)` and
`-(1)/(2)`

Explanation:

`h(t)=-16t^2+16t+12`

1) Find the x-coordinate of the vertex

`x=(-b)/(2a)` where the equation is
`f(x)=ax^2+bx+c`

`x=(-16)/(2(-16)) \\x=(-16)/(-32) \\x=(16)/(32) \\x=(1)/(2)`

2) Find the y-coordinate of the vertex

Plug the x-coordinate
`(1)/(2)` back into the original equation

`h((1)/(2) )=-16((1)/(2)) ^2+16((1)/(2))+12\\h((1)/(2) )=-16((1)/(4))+8+12\\h((1)/(2) )=-4+20\\h((1)/(2) )=16`

Therefore, the vertex of the parabola is
`((1)/(2) ,16)`.

3) Find the zeros of the parabola

Rewrite the equation so that it equals 0

`0=-16t^2+16t+12`

Divide both sides by -16

`0=t^2-t-(12)/(16)`

Simplify the fraction

`0=t^2-t-(3)/(4)`

Complete the square; first isolate t²-t

`t^2-t=(3)/(4)`

Complete the square; Add
`((1)/(2) )^2` to both sides

`t^2-t+((1)/(2))^2=(3)/(4)+((1)/(2))^2\\(t-(1)/(2))^2 =(3)/(4)+(1)/(4) \\(t-(1)/(2))^2=1\\(t-(1)/(2))^2=1`

Take the square root of both sides

`t-(1)/(2) =\±√(1)\\t-(1)/(2) =\±1`

Isolate t

`t=(1)/(2)\±1`

Find the zeros

`t=(1)/(2)+1\\t=(3)/(2)\\or\\t=(1)/(2)-1\\t=-(1)/(2)`

Therefore, the zeros of the parabola are
`(3)/(2)` and
`-(1)/(2)`.

I hope this helps!