Question

How to graph a parabola #h(t)=-16t^2+280t+17?

Answer 1

`h(t) = -16t^2+280t+17` is a very steep parabola with vertex at `(8.75, 1242)` and intersections with the `t` axis at approximately `(-0.06, 0)` and `(17.56, 0)`

Explanation:

`h(t) = -16t^2+280t+17`

In general, we have:

`at^2+bt+c = a(t+b/(2a))^2 + (c-b^2/(4a))`

The vertex of this parabola is `(-b/2a, c-b^2/(4a))`

The axis of symmetry is `t = -b/(2a)`

The intersections (if any) with the `t` axis are at:

`((-b +- sqrt(b^2-4ac))/(2a), 0)`

The intersection with the vertical axis is at:

`(0, c)`

In our case, we have `a=-16`, `b=280` and `c=17`

The discriminant `Delta` is given by the formula:

`Delta = b^2-4ac = 280^2-(4xx-16xx17) =79488`

`= 2^7*3^3*23 = 24^2*138`

Being positive, but not a perfect square, the parabola does intersect the `t` axis at the points:

`((280+-24sqrt(138))/32, 0) = ((35+-3sqrt(138))/4, 0)`

That is approximately `(-0.06, 0)` and `(17.56, 0)`

The axis of symmetry is `x = 35/4`.

The vertex is at `(35/4, 17+35^2) = (8.75, 1242)`

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 `h(x/100)`

graph{-16(x/100)^2+280(x/100)+17 [-2128, 3710, -1520, 1400]}