Find the equation that has 2, 1, and -1 as roots
1 answer:
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If you are asking what is the graph of y = 3x^2 -2x+1.
Then, the attached file would be the answer.
To check, b^2 - 4(a)(c), for each equation and use these facts:
If b^2 - 4(a)(c) = 0, there is only one real root meaning, the graph touches the x-axis only in one point.
If b^2 - 4ac > 0, there are two real roots meaning, the graph touches the x-axis in two different points.
If b2 - 4ac < 0, there are no real roots then the graph does not touch the x-axis. This would be the case for y = 3x^2 - 2x + 1.
Solution:
(-2)^2 -4(3)(1) = 4 - 12 = -8 < 0 will result in not real roots.
Then, the attached file would be the answer.
To check, b^2 - 4(a)(c), for each equation and use these facts:
If b^2 - 4(a)(c) = 0, there is only one real root meaning, the graph touches the x-axis only in one point.
If b^2 - 4ac > 0, there are two real roots meaning, the graph touches the x-axis in two different points.
If b2 - 4ac < 0, there are no real roots then the graph does not touch the x-axis. This would be the case for y = 3x^2 - 2x + 1.
Solution:
(-2)^2 -4(3)(1) = 4 - 12 = -8 < 0 will result in not real roots.