Answer:
Vertex: [tex](\frac{1}{2} ,16)[/tex]
Zeros: [tex]\frac{3}{2}[/tex] and [tex]-\frac{1}{2}[/tex]
Step-by-step explanation:
[tex]h(t)=-16t^2+16t+12[/tex]
1) Find the x-coordinate of the vertex
[tex]x=\frac{-b}{2a}[/tex] where the equation is [tex]f(x)=ax^2+bx+c[/tex]
[tex]x=\frac{-16}{2(-16)} \\x=\frac{-16}{-32} \\x=\frac{16}{32} \\x=\frac{1}{2}[/tex]
2) Find the y-coordinate of the vertex
Plug the x-coordinate [tex]\frac{1}{2}[/tex] back into the original equation
[tex]h(\frac{1}{2} )=-16(\frac{1}{2}) ^2+16(\frac{1}{2})+12\\h(\frac{1}{2} )=-16(\frac{1}{4})+8+12\\h(\frac{1}{2} )=-4+20\\h(\frac{1}{2} )=16[/tex]
Therefore, the vertex of the parabola is [tex](\frac{1}{2} ,16)[/tex].
3) Find the zeros of the parabola
Rewrite the equation so that it equals 0
[tex]0=-16t^2+16t+12[/tex]
Divide both sides by -16
[tex]0=t^2-t-\frac{12}{16}[/tex]
Simplify the fraction
[tex]0=t^2-t-\frac{3}{4}[/tex]
Complete the square; first isolate t²-t
[tex]t^2-t=\frac{3}{4}[/tex]
Complete the square; Add [tex](\frac{1}{2} )^2[/tex] to both sides
[tex]t^2-t+(\frac{1}{2})^2=\frac{3}{4}+(\frac{1}{2})^2\\(t-\frac{1}{2})^2 =\frac{3}{4}+\frac{1}{4} \\(t-\frac{1}{2})^2=1\\(t-\frac{1}{2})^2=1[/tex]
Take the square root of both sides
[tex]t-\frac{1}{2} =\±\sqrt{1}\\t-\frac{1}{2} =\±1[/tex]
Isolate t
[tex]t=\frac{1}{2}\±1[/tex]
Find the zeros
[tex]t=\frac{1}{2}+1\\t=\frac{3}{2}\\or\\t=\frac{1}{2}-1\\t=-\frac{1}{2}[/tex]
Therefore, the zeros of the parabola are [tex]\frac{3}{2}[/tex] and [tex]-\frac{1}{2}[/tex].
I hope this helps!
