The zeros of the function f(x)=(x+2)^2 - 25 are?

2 answers:

5 0

Zeros of the function
f(x) = (x + 2)² - 25
f(x) = (x + 2)(x + 2) - 25
f(x) = x(x + 2) + 2(x + 2) - 25
f(x) = x(x) + x(2) + 2(x) + 2(2) - 25
f(x) = x² + 2x + 2x + 4 - 25
f(x) = x² + 4x + 4 - 25
f(x) = x² + 4x - 21
x² + 4x - 21 = 0
x = -(4) +/- √((4)² - 4(1)(-21))
 2(1)
x = -4 +/- √(16 + 84)
 2
x = -4 +/- √(100)
 2
x = -4 +/- 10
 2
x = -2 + 5
x = -2 + 5 x = -2 - 5
x = 3 x = -7
f(x) = x² + 4x - 21
f(3) = (3)² + 4(3) - 21
f(3) = 9 + 12 - 21
f(3) = 21 - 21
f(3) = 0
(x, f(x)) = (3, 0)
or
f(x) = x² + 4x - 21
f(-7) = (-7)² + 4(-7) - 21
f(-7) = 49 - 28 - 21
f(-7) = 21 - 21
f(-7) = 0
(x, f(x)) = (-7, 0)

Vertex
X - Intercept
-b/2a = -(4)/2(1) = -4/2 = -2

Y - Intercept
y = x² + 4x - 21
y = (-2)² + 4(-2) - 21
y = 4 - 8 - 21
y = -4 - 21
y = -25
(x, y) = (-2, -25)

Serjik [45]3 years ago

5 0

Y=x^2+4x+4-25
y=x^2+4x-21
y=(x+7)(x-3)
zeros are -7 and 3

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Alex73 [517]

Answer:

Step-by-step explanation:

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The equation we will use is the one for displacement:

Δx = $v_0t+\frac{1}{2}at^2$ and filling in:

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Now we use that time in the x-dimension. Here's what we know in that dimension specifically:

a = 0 (acceleration in this dimension is always 0)

v₀ = 80 ft/sec

t = .968 sec

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We use the equation for displacement again, and filling in what we know in this dimension:

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3 years ago

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Goshia [24]

Answer:

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