Given the graphs of g(x) = x^2 + 3x + 2 and 2y + mx + 8 = 0. Find m such that the graphs have
one intersection
no intersection
Would I need to set the equations equal to each other for both?
year 11 2017
year 12 2018
year boredsatan 2019
year LOL 2020-infinity
Last edited by Drongoski; 15 May 2017 at 11:00 PM.
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Yes. In the first one, since there is a 2/(x+2)^2, this can only take on positive values (and moreover can take on any positive value), because of the square in the denominator (and positive constant numerator). This causes the range to become (2, infinity).
With the second one, the 2/(x+2) can take on any non-zero real value (including negative ones), and can't take the value 0. This makes the range be what you wrote (ℝ \ {2}).
Bump!
How would you express y = 2x^2 - 8x + 12 in turning point form
Is this working right?
2(x^2 - 4x + 6)
x^2 - 4x + 4 - 4 + 6
(x-2)^2 + 2
2(x-2)^2+2
year 11 2017
year 12 2018
year boredsatan 2019
year LOL 2020-infinity
Last edited by Drongoski; 16 May 2017 at 9:21 PM.
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y = 0.5x + 5
3y = √(3) x + 15
are these the same?
year 11 2017
year 12 2018
year boredsatan 2019
year LOL 2020-infinity
find the equation of a line which makes an angle of 30 degrees with the positive x axis and cuts the y axis at 5
year 11 2017
year 12 2018
year boredsatan 2019
year LOL 2020-infinity
Then this line makes an angle of 30 deg with the positive x-axis. Its gradient is tan(30 deg) = 1/sqrt(3)
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IB: Maths Studies, Maths SL & Maths HL; HSC: 2U, 3U & 4U
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