How to graph a parabola #h(t)=-16t^2+280t+17?
1 Answer
Explanation:
In general, we have:
The vertex of this parabola is
The axis of symmetry is
The intersections (if any) with the
The intersection with the vertical axis is at:
In our case, we have
The discriminant
Being positive, but not a perfect square, the parabola does intersect the
That is approximately
The axis of symmetry is
The vertex is at
This is parabola is very 'steep', so it will help to have a very different scale on the horizontal and vertical axes.
Here I have graphed
graph{-16(x/100)^2+280(x/100)+17 [-2128, 3710, -1520, 1400]}