Write y = (x – 6)(x – 4)(x + 2) in standard form. A. y = x3 + 48 B. y = x3 – 8x2 + 4x + 48 C. y = x3 – 2x2 + 4x – 8 D. y = x3 – 8x2 – 8x + 24
y = (x – 6)(x – 4)(x + 2) in standard form is y = x^3 – 8x^2 + 4x + 48 (x – 6)(x – 4)(x + 2) = (x^2 - 10x + 24)(x + 2) = x^3 - 8x^2 + 4x + 48
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Asked 5/18/2014 10:35:56 AM
Updated 5/18/2014 10:56:01 AM
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3
y = (x – 6)(x – 4)(x + 2) in standard form is y = x^3 – 8x^2 + 4x + 48

(x – 6)(x – 4)(x + 2) = (x^2 - 10x + 24)(x + 2) = x^3 - 8x^2 + 4x + 48
Added 5/18/2014 10:56:01 AM
This answer has been confirmed as correct and helpful.
Confirmed by alfred123 [5/18/2014 10:58:06 AM]

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The minimum value of a parabola that opens upward will be its vertex. A. True B. False
Weegy: The vertex of a parabola is the point where the parabola crosses its axis. If the coefficient of the x2 term is positive, the vertex will be the lowest point on the graph, the point at the bottom of the ?U?-shape. [ If the coefficient of the x2 term is negative, the vertex will be the highest point on the graph, the point at the top of the ?U?-shape. The standard equation of a parabola is y = ax2 + bx + c. But the equation for a parabola can also be written in "vertex form": y = a(x ? h)2 + k ] (More)
Question
Updated 5/19/2014 8:47:34 PM
The minimum value of a parabola that opens upward will be its vertex. TRUE.
Added 5/19/2014 8:47:34 PM
This answer has been confirmed as correct and helpful.
Confirmed by jeifunk [5/19/2014 8:54:03 PM]
. Identify the vertex and y-intercept of the graph of f(x) = –3(x + 6)2 + 9. A. (–6, –9), y-intercept –99 B. (6, 9), y-intercept –9 C. (–6, 9), y-intercept –99 D. (6, –9), y-intercept –9
Question
Updated 5/19/2014 8:45:48 PM
The vertex and y-intercept of the graph of f(x) = –3(x + 6)^2 + 9 is vertex (–6, 9), y-intercept –99.

Added 5/19/2014 8:45:48 PM
This answer has been confirmed as correct and helpful.
Confirmed by jeifunk [5/19/2014 8:54:09 PM]
Identify the vertex and y-intercept of the graph of f(x) = –4(x + 3)2 + 7. A. (3, –7), y-intercept –5 B. (–3, 7), y-intercept –29 C. (–3, –7), y-intercept –29 D. (3, 7), y-intercept –5
Question
Updated 5/19/2014 8:44:22 PM
The vertex and y-intercept of the graph of f(x) = –4(x + 3)^2 + 7 is vertex (–3, 7), y-intercept –29.
Added 5/19/2014 8:44:22 PM
This answer has been confirmed as correct and helpful.
Confirmed by jeifunk [5/19/2014 8:45:10 PM]
The vertex of f(x) = 3(x+2)2 – 4 is (-2,-4). A. True B. False
Weegy: -2.5 + 6.3 + (-4.1) = -0.3 (More)
Question
Updated 5/19/2014 8:06:07 PM
The vertex of f(x) = 3(x+2)^2 – 4 is (-2, -4). TRUE.
Added 5/19/2014 8:06:07 PM
This answer has been confirmed as correct and helpful.
Confirmed by jeifunk [5/19/2014 8:23:02 PM]
Factor 4y2 – 9. A. (2y + 3)(2y + 3) B. (2y - 3)(y - 3) C. (y + 3)(2y - 3) D. (2y + 3)(2y - 3)
Weegy: (4y^2)(2y) = 8y^3 (More)
Question
Updated 5/15/2014 11:26:03 PM
4y^2 - 9 = (2y + 3)(2y - 3)
Added 5/15/2014 11:26:03 PM
This answer has been confirmed as correct and helpful.
Confirmed by jeifunk [5/16/2014 12:27:25 AM]
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