PLEASE HELP The y-intercept of the equation y = 12x - 8 is 8. True False
2 answers:
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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.
Answer:
Step-by-step explanation:
Easy way to do this is step by step. Your quadratic, from your entry, must be
.
Step by step looks like this, one thing at a time:
becomes
becomes
and this of course is
Do the same with the subtraction sign to get the other solution.
If you're unsure of how to enter it into your calculator, do it step by step so you don't mess up the sign. If you enter it incorrectly, you could end up with an imaginary number when it should be real, or a real one that should be imaginary.
Just my advice as a high school math teacher.