How do you find the zeros of # y = -x^2 +32x + 19 # using the quadratic formula?

Answer 1

See a solution process below:

The quadratic formula states:

For #color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0#, the values of #x# which are the solutions to the equation are given by:
#x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))#

Substituting:

#color(red)(-1)# for #color(red)(a)#
#color(blue)(32)# for #color(blue)(b)#
#color(green)(19)# for #color(green)(c)# gives:
#x = (-color(blue)(32) +- sqrt(color(blue)(32)^2 - (4 * color(red)(-1) * color(green)(19))))/(2 * color(red)(-1))#
#x = (-color(blue)(32) +- sqrt(1024 - (-76)))/-2#
#x = (-color(blue)(32) +- sqrt(1024 + 76))/-2#
#x = (-color(blue)(32) +- sqrt(1100))/-2#
#x = (-color(blue)(32) - sqrt(100 * 11))/-2# and #x = (-color(blue)(32) + sqrt(100 * 11))/-2#
#x = (-color(blue)(32) - sqrt(100)sqrt(11))/-2# and #x = (-color(blue)(32) + sqrt(100)sqrt(11))/-2#
#x = (-color(blue)(32) - 10sqrt(11))/-2# and #x = (-color(blue)(32) + 10sqrt(11))/-2#
#x = (-color(blue)(32))/-2 - (10sqrt(11))/-2# and #x = (-color(blue)(32))/-2 + (10sqrt(11))/-2#
#x = 16 + 5sqrt(11)# and #x = 16 - 5sqrt(11)#
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Answer 2

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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Answer from HIX Tutor

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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