How do you find the zeroes for #y=9x^2+6x-1#?
-0.805 or 0.138
The quadratic formula would then appear as follows:
graph{[-5.285, 4.58, -2.74, 2.193]} 9x^2+6x-1
:-)
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To find the zeroes of the quadratic equation ( y = 9x^2 + 6x - 1 ), you can use the quadratic formula ( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} ), where ( a = 9 ), ( b = 6 ), and ( c = -1 ). Plugging in these values, you get:
( x = \frac{{-6 \pm \sqrt{{6^2 - 4(9)(-1)}}}}{{2(9)}} )
Simplify under the square root:
( x = \frac{{-6 \pm \sqrt{{36 + 36}}}}{{18}} ) ( x = \frac{{-6 \pm \sqrt{{72}}}}{{18}} ) ( x = \frac{{-6 \pm 6\sqrt{2}}}{{18}} ) ( x = \frac{{-1 \pm \sqrt{2}}}{{3}} )
Therefore, the zeroes of the equation are ( x = \frac{{-1 + \sqrt{2}}}{{3}} ) and ( x = \frac{{-1 - \sqrt{2}}}{{3}} ).
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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