# 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]}