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)#

#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]}

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Answer 2

To graph the parabola ( h(t) = -16t^2 + 280t + 17 ), follow these steps:

  1. Identify the vertex using the formula ( t = \frac{-b}{2a} ), where ( a = -16 ) and ( b = 280 ).
  2. Substitute the value of ( t ) found in step 1 into the equation to find the corresponding value of ( h ).
  3. Identify the y-intercept by setting ( t = 0 ) and solving for ( h ).
  4. Determine the x-intercepts by setting ( h = 0 ) and solving for ( t ).
  5. Plot the vertex, y-intercept, and x-intercepts on the coordinate plane.
  6. Use the symmetry of the parabola to plot additional points if needed.
  7. Sketch the parabola passing through the plotted points.
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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