Hi,
I have the following quadratic function: -x^2-4x+4.
After completing the square, this quadratic function can be expressed in the standard form like so:
-(x+2)^2+8
What are the exact steps I would take to find the x-intercepts of this parabola? Obviously, I'd set -(x+2)^2+8 = 0, but from there I've been getting a bit stuck. I have subtracted 8 from both sides of the equation, multiplied both sides by -1, yielding (x+2)^2 = 8. I took the square root of both sides of the equation, then subtract 2, such that it would appear that x = √8 - 2. Is this correct? Thanks.
I have the following quadratic function: -x^2-4x+4.
After completing the square, this quadratic function can be expressed in the standard form like so:
-(x+2)^2+8
What are the exact steps I would take to find the x-intercepts of this parabola? Obviously, I'd set -(x+2)^2+8 = 0, but from there I've been getting a bit stuck. I have subtracted 8 from both sides of the equation, multiplied both sides by -1, yielding (x+2)^2 = 8. I took the square root of both sides of the equation, then subtract 2, such that it would appear that x = √8 - 2. Is this correct? Thanks.
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(x+2)^2 = 8
x = +√8 - 2 or x = -√8 - 2
x = +√8 - 2 or x = -√8 - 2
-
f(x) = -x² - 4x + 4
f(x) = -(x² + 4x .....) + 4
f(x) = -(x² + 4x + 4) + 4 + 4
f(x) = -(x + 2)² + 8
so vertex is (-2,8) and the parabola opens down.
solving
-(x + 2)² + 8 = 0
(x + 2)² = 8
x + 2 = ±2√2
x = -2 ± 2√2
f(x) = -(x² + 4x .....) + 4
f(x) = -(x² + 4x + 4) + 4 + 4
f(x) = -(x + 2)² + 8
so vertex is (-2,8) and the parabola opens down.
solving
-(x + 2)² + 8 = 0
(x + 2)² = 8
x + 2 = ±2√2
x = -2 ± 2√2
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the product of three negative numbers is positive. give a example.