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For the parabola in the xy-plane, find the following. 2y =x −

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For the parabola in the xy-plane, find the following. 2y =x − [#permalink] New post 24 May 2019, 08:15
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For the parabola in the xy-plane, find the following. \(y =x^2 −4x−12\)

(a) The x-intercepts

(b) The y-intercept

(c) Coordinates of the vertex


[Reveal] Spoiler: OA
(a) and x=−2 x=6 (b) y=−12 (c) (2,−16)


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Re: For the parabola in the xy-plane, find the following. 2y =x − [#permalink] New post 28 May 2019, 06:26
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Carcass wrote:
For the parabola in the xy-plane, find the following. \(y =x^2 −4x−12\)

(a) The x-intercepts

(b) The y-intercept

(c) Coordinates of the vertex

[Reveal] Spoiler: OA
(a) and x=−2 x=6 (b) y=−12 (c) (2,−16)




(a) The x-intercepts
The x-intercepts are points where the y-coordinate is ZERO.
So, replace y with 0 to get: \(0=x^2−4x−12\)
Factor to get: \(0=(x-6)(x+2)\)
So, either x = 6 or x = -2
So, the x-intercepts are 6 and -2


(b) The y-intercept
The y-intercept is the point where the x-coordinate is ZERO.
So, replace x with 0 to get: \(y=0^2−4(0)−12\)
Evaluate: \(y=−12\)
So, the y-intercept is -12


(c) Coordinates of the vertex
Take: y = x² − 4x − 12
Rewrite as: y = x² − 4x + 4 − 12 - 4 [the net effect of adding 4 and subtracting 4 is ZERO]
Rewrite as: y = (x² − 4x + 4) − 16
Factor to get: y = (x - 2)² − 16
When we compare this to the graph of the BASE equation y = x², we can see that it has been shifted 2 units to the RIGHT, and 16 units DOWN.
So, the vertex is now at (2, -16)

Cheers,
Brent
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Re: For the parabola in the xy-plane, find the following. 2y =x −   [#permalink] 28 May 2019, 06:26
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