# How do you find the zeros, real and imaginary, of #y=-4x^2+4x-16# using the quadratic formula?

2 imaginary roots:

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To find the zeros of the quadratic function ( y = -4x^2 + 4x - 16 ) using the quadratic formula, first identify the coefficients: ( a = -4 ), ( b = 4 ), and ( c = -16 ).

Then, substitute these values into the quadratic formula:

[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} ]

Substitute the coefficients into the formula:

[ x = \frac{{-4 \pm \sqrt{{4^2 - 4(-4)(-16)}}}}{{2(-4)}} ]

Simplify inside the square root:

[ x = \frac{{-4 \pm \sqrt{{16 - 256}}}}{{-8}} ] [ x = \frac{{-4 \pm \sqrt{{-240}}}}{{-8}} ]

Since the discriminant ( b^2 - 4ac = -240 ) is negative, the solutions will involve imaginary numbers.

[ x = \frac{{-4 \pm \sqrt{{240i}}}}{{-8}} ]

Further simplification involves taking the square root of ( i ):

[ x = \frac{{-4 \pm 4\sqrt{{15i}}}}{{-8}} ] [ x = \frac{{1 \pm \sqrt{{15i}}}}{2} ]

Thus, the zeros of the function ( y = -4x^2 + 4x - 16 ) are complex numbers: ( \frac{{1 + \sqrt{{15i}}}}{2} ) and ( \frac{{1 - \sqrt{{15i}}}}{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.

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