How do you find the zeros of # y = -x^2 +32x + 19 # using the quadratic formula?
See a solution process below:
The quadratic formula states:
Substituting:
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To find the zeros of the quadratic function (y = -x^2 + 32x + 19) using the quadratic formula, first identify the coefficients (a), (b), and (c) from the standard form of the quadratic equation, (y = ax^2 + bx + c). In this case, (a = -1), (b = 32), and (c = 19).
Then, substitute these values into the quadratic formula:
[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}]
Substitute (a = -1), (b = 32), and (c = 19) into the formula:
[x = \frac{{-32 \pm \sqrt{{32^2 - 4(-1)(19)}}}}{{2(-1)}}]
Simplify under the square root:
[x = \frac{{-32 \pm \sqrt{{1024 + 76}}}}{{-2}}]
[x = \frac{{-32 \pm \sqrt{{1100}}}}{{-2}}]
[x = \frac{{-32 \pm 10\sqrt{{11}}}}{{-2}}]
Finally, simplify further to get the zeros:
[x = \frac{{32 \pm 10\sqrt{{11}}}}{2}]
Therefore, the zeros of the quadratic function (y = -x^2 + 32x + 19) are:
[x = \frac{{32 + 10\sqrt{{11}}}}{2}]
[x = \frac{{32 - 10\sqrt{{11}}}}{2}]
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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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