Which equation could generate the curve in the graph below?y=3x^2-2x+1
1 answer:
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.
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Answer:
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Step-by-step explanation:
If a function is bijective and 1-to-1, then it will have an inverse function. Consequentially, they will be symmetrical about the line , which is a diagonal line passing through the origin at a 45 degree angle.
None of the graphs look correct though, but it also seems that some options are cut out, so make sure to choose the correct graph given the characteristics I've previously described.