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

1 Answer
Jun 20, 2015

#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]}