# Thread: VCE Maths questions help

1. ## Re: VCE Maths questions help

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?

2. ## Re: VCE Maths questions help

Originally Posted by 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?
Essentially.

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

3. ## Re: VCE Maths questions help

Originally Posted by InteGrand
Essentially.

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

4. ## Re: VCE Maths questions help

Originally Posted by boredsatan
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.

$]2, \infty [ meaning y > 2$

5. ## Re: VCE Maths questions help

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

$]2, \infty [ meaning y > 2$
Fairly sure they use the parentheses in VCE.

6. ## Re: VCE Maths questions help

Originally Posted by leehuan
Fairly sure they use the parentheses in VCE.
Yes, parentheses and brackets are used in vce

7. ## Re: VCE Maths questions help

Originally Posted by 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} ?

8. ## Re: VCE Maths questions help

Originally Posted by boredsatan
So would the range of y = 2/(x+2) + 2 be R {2} ?
ℝ \ {2} (You forgot a backslash.)

9. ## Re: VCE Maths questions help

Originally Posted by InteGrand
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}

10. ## Re: VCE Maths questions help

Originally Posted by boredsatan
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}).

11. ## Re: VCE Maths questions help

Originally Posted by InteGrand
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}?

12. ## Re: VCE Maths questions help

Originally Posted by boredsatan
Although the domain for both would be R\{2}?
Correct!

13. ## Re: VCE Maths questions help

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

14. ## Re: VCE Maths questions help

Originally Posted by 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
That's correct. So

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

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

15. ## Re: VCE Maths questions help

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

16. ## Re: VCE Maths questions help

Originally Posted by boredsatan
y = 0.5x + 5
3y = √(3) x + 15
are these the same?
No

17. ## Re: VCE Maths questions help

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

18. ## Re: VCE Maths questions help

Originally Posted by 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?

19. ## Re: VCE Maths questions help

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

20. ## Re: VCE Maths questions help

Originally Posted by Drongoski
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 ?

21. ## Re: VCE Maths questions help

Originally Posted by boredsatan
is the answer 1/sqrt(3) x + 5 ?
That isn't an equation.

22. ## Re: VCE Maths questions help

Originally Posted by 1729
That isn't an equation.
y = 1/sqrt(3) x + 5 ?

23. ## Re: VCE Maths questions help

Originally Posted by boredsatan
y = 1/sqrt(3) x + 5 ?
Correct!

24. ## Re: VCE Maths questions help

Originally Posted by InteGrand
No
y = 1/sqrt(3) x + 5
3y = √(3) x + 15
are these the same?

25. ## Re: VCE Maths questions help

Originally Posted by boredsatan
y = 1/sqrt(3) x + 5
3y = √(3) x + 15
are these the same?
Yes

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