How do you graph the parabola #h(t)=-16t^2+280t+17 # using vertex, intercepts and additional points?
Graph parabola
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To graph the parabola h(t) = -16t^2 + 280t + 17, you can start by finding the vertex, intercepts, and additional points.
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Vertex: The vertex of a parabola in the form h(t) = at^2 + bt + c is given by the point (h, k), where h = -b/(2a) and k = h(h).
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Intercepts: To find the y-intercept, set t = 0 and solve for h(0). To find the x-intercepts, set h(t) = 0 and solve for t.
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Additional Points: Choose additional values of t to evaluate h(t) and plot those points on the graph.
Once you have these points, you can plot them on a graph and sketch the parabola passing through these 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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