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what would be a suitable table of values for this equation? y=x^2+3x

this is what I have for x -3, -2, -1, 0, 1

2006-11-19 06:23:25 · 2 answers · asked by George F 1 in Science & Mathematics Mathematics

2 answers

Very ingenious to have a "That's Me" account. You can give yourself ten points over and over.

for x = -3, x^2 = 9 and 3x = -9. Result = 0
for x = -2, x^2 = 4 and 3x = -6. Result = -2
for x = -1, x^2 = 1 and 3x = -3. Result = -2
for x = 0, x^2 = 0 and 3x = 0. Result = 0
for x = 1, x^2 = 1 and 3x = 3. Result = 3

Only your zero value is correct.

This is an upward-opening parabola with its minimum point at

x = -1.5

It crosses the x-axis at (-3, 0) and (0, 0). Obviously, it crosses the y-axis at (0, 0) also.

2006-11-19 06:32:33 · answer #1 · answered by ? 6 · 0 0

-3 = 3
-2 = 1
-1 = -1
0 = 0
1 = 2

2006-11-19 06:29:26 · answer #2 · answered by ThatsMe!!!! 1 · 0 0

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