How to graph a parabola #h(t)=-16t^2+280t+17?
In general, we have:
The intersection with the vertical axis is at:
This is parabola is very 'steep', so it will help to have a very different scale on the horizontal and vertical axes.
graph{-16(x/100)^2+280(x/100)+17 [-2128, 3710, -1520, 1400]}
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To graph the parabola ( h(t) = -16t^2 + 280t + 17 ), follow these steps:
- Identify the vertex using the formula ( t = \frac{-b}{2a} ), where ( a = -16 ) and ( b = 280 ).
- Substitute the value of ( t ) found in step 1 into the equation to find the corresponding value of ( h ).
- Identify the y-intercept by setting ( t = 0 ) and solving for ( h ).
- Determine the x-intercepts by setting ( h = 0 ) and solving for ( t ).
- Plot the vertex, y-intercept, and x-intercepts on the coordinate plane.
- Use the symmetry of the parabola to plot additional points if needed.
- Sketch the parabola passing through the plotted points.
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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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