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How do you accurately graph y -x2 8x 5 as a parabola? - Answers

There are two (main) ways to graph this parabola. The first is to simply substitute in values of x, find the corresponding y values and then plot those points and connect the points with a curve. The second way is to 'complete the square' so that we can find where the maximum or minimum occurs and the y-intercept and graph from those two points. This is the process described below. y = -x2 + 8x + 5 y = -(x2 - 8x) + 5 y = -(x2 - 8x + 16 - 16) + 5 y = -(x2 - 8x + 16) + 16 + 5 y = -(x-4)2 + 21 From this we can see that a maximum occurs at (4,21). To find the y-intercept, substitute in x=0, giving y=5. From these two points, we get a pretty good idea of what the graph looks like. If you want more accuracy, substitute in more values of x.



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How do you accurately graph y -x2 8x 5 as a parabola? - Answers

https://math.answers.com/math-and-arithmetic/How_do_you_accurately_graph_y_-x2_8x_5_as_a_parabola

There are two (main) ways to graph this parabola. The first is to simply substitute in values of x, find the corresponding y values and then plot those points and connect the points with a curve. The second way is to 'complete the square' so that we can find where the maximum or minimum occurs and the y-intercept and graph from those two points. This is the process described below. y = -x2 + 8x + 5 y = -(x2 - 8x) + 5 y = -(x2 - 8x + 16 - 16) + 5 y = -(x2 - 8x + 16) + 16 + 5 y = -(x-4)2 + 21 From this we can see that a maximum occurs at (4,21). To find the y-intercept, substitute in x=0, giving y=5. From these two points, we get a pretty good idea of what the graph looks like. If you want more accuracy, substitute in more values of x.



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https://math.answers.com/math-and-arithmetic/How_do_you_accurately_graph_y_-x2_8x_5_as_a_parabola

How do you accurately graph y -x2 8x 5 as a parabola? - Answers

There are two (main) ways to graph this parabola. The first is to simply substitute in values of x, find the corresponding y values and then plot those points and connect the points with a curve. The second way is to 'complete the square' so that we can find where the maximum or minimum occurs and the y-intercept and graph from those two points. This is the process described below. y = -x2 + 8x + 5 y = -(x2 - 8x) + 5 y = -(x2 - 8x + 16 - 16) + 5 y = -(x2 - 8x + 16) + 16 + 5 y = -(x-4)2 + 21 From this we can see that a maximum occurs at (4,21). To find the y-intercept, substitute in x=0, giving y=5. From these two points, we get a pretty good idea of what the graph looks like. If you want more accuracy, substitute in more values of x.

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      There are two (main) ways to graph this parabola. The first is to simply substitute in values of x, find the corresponding y values and then plot those points and connect the points with a curve. The second way is to 'complete the square' so that we can find where the maximum or minimum occurs and the y-intercept and graph from those two points. This is the process described below. y = -x2 + 8x + 5 y = -(x2 - 8x) + 5 y = -(x2 - 8x + 16 - 16) + 5 y = -(x2 - 8x + 16) + 16 + 5 y = -(x-4)2 + 21 From this we can see that a maximum occurs at (4,21). To find the y-intercept, substitute in x=0, giving y=5. From these two points, we get a pretty good idea of what the graph looks like. If you want more accuracy, substitute in more values of x.

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