VCE Maths questions help (2 Viewers)

[boredsatan]

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?

 

[InteGrand]

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?

Essentially.

(Make sure to first express y in terms of x for the 2y + mx + 8 = 0.)

 

[boredsatan]

Essentially.

(Make sure to first express y in terms of x for the 2y + mx + 8 = 0.)

Thanks for claryfying

 

[Drongoski]

could it be (2,infinity), even though the graph never touches 2?

That's the range in the "old" notation. Maybe for VCE, like for the IB, you use ]2, infinity symbol [, i.e.

 

[leehuan]

That's the range in the "old" notation. Maybe for VCE, like for the IB, you use ]2, infinity symbol [, i.e.

Fairly sure they use the parentheses in VCE.

 

[boredsatan]

Fairly sure they use the parentheses in VCE.

Yes, parentheses and brackets are used in vce

 

[boredsatan]

could it be (2,infinity), even though the graph never touches 2?

So would the range of y = 2/(x+2) + 2 be R {2} ?

 

[InteGrand]

So would the range of y = 2/(x+2) + 2 be R {2} ?

ℝ \ {2} (You forgot a backslash.)

 

[boredsatan]

ℝ \ {2} (You forgot a backslash.)

Is there a reason why for y = 2/(x+2)^2 + 2, the range is (2, infinity),
but for y = 2/(x+2) + 2, the range is R\ {2}

 

[InteGrand]

Is there a reason why for y = 2/(x+2)^2 + 2, the range is (2, infinity),
but for y = 2/(x+2) + 2, the range is R\ {2}

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}).

 

[boredsatan]

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}).

Although the domain for both would be R\{2}?

 

[InteGrand]

Although the domain for both would be R\{2}?

Correct!

 

[boredsatan]

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

 

[Drongoski]

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

That's correct. So

y - 2 = 2 (x - 2)^2

So the turning point (equivalently the vertex of the parabola) is (2,2))

 

[boredsatan]

y = 0.5x + 5
3y = √(3) x + 15
are these the same?

 

[InteGrand]

y = 0.5x + 5
3y = √(3) x + 15
are these the same?

No

 

[boredsatan]

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

 

[boredsatan]

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

Anyone?

 

[Drongoski]

Then this line makes an angle of 30 deg with the positive x-axis. Its gradient is tan(30 deg) = 1/sqrt(3)

 

[boredsatan]

Then this line makes an angle of 30 deg with the positive x-axis. Its gradient is tan(30 deg) = 1/sqrt(3)

is the answer 1/sqrt(3) x + 5 ?