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prove that the graph of y=ay=ax^{3}+bx^{2}+cx+has two distinct turning points if b^{2}> 3ac
geometrical applications (1 Viewer)
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D94
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Find the first derivative and let it equal zero to find the stationary points: y' = 3ax^2 + 2bx + c = 0
So we need to get two x coordinates, and the best way is to use the quadratic formula:c}}{2(3a)} \\ x = \frac{-2b +/- 2\sqrt{b^{2} - 3ac}}{6a} \\Now,\ to\ have\ two\ distinct\ roots,\ underneath\ the\ square\ root\ must\ be\ greater\ than\ zero. \\\therefore b^{2} - 3ac > 0 \\\therefore b^{2} > 3ac)
Now since
, there are two x coordinates; one with the positive sign and one with the negative sign.
So we need to get two x coordinates, and the best way is to use the quadratic formula:
Now since