Question 54179
{{{h(t)=-16t^2+256}}}
{{{t^2}}} tells you that this is a parabola.
The - infront of 16, tells you that this parabola is upside down, so it's vertex would be the highest point of the graph.
To find the x value of the vertex of a quadratic equation that is in standard form {{{highlight(ax^2+bx+c)}}}, we use the formula:
{{{highlight(x=-b/2a)}}}
In your case there is no bx, so b=0.
{{{x=-0/2(-16)}}}
{{{x=0}}}
Substitute that value x=0 into your equation to find y (or h(t)).
{{{h(0)=-16(0)^2+256}}}
{{{h(0)=0+256}}}
{{{h(0)=256}}}
The highest point is at (0,256).
{{{graph(300,200,-20,20,100,270,-16x^2+256)}}}
Happy Calculating!!!