# 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?  Reply With Quote

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.)  Reply With Quote

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  Reply With Quote

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.  Reply With Quote

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.

Fairly sure they use the parentheses in VCE.  Reply With Quote

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  Reply With Quote

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} ?  Reply With Quote

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.)  Reply With Quote

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}  Reply With Quote

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}).  Reply With Quote

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}?  Reply With Quote

12. ## Re: VCE Maths questions help Originally Posted by boredsatan Although the domain for both would be R\{2}?
Correct!  Reply With Quote

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  Reply With Quote

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))  Reply With Quote

15. ## Re: VCE Maths questions help

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

16. ## Re: VCE Maths questions help Originally Posted by boredsatan y = 0.5x + 5
3y = √(3) x + 15
are these the same?
No  Reply With Quote

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  Reply With Quote

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?  Reply With Quote

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)  Reply With Quote

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 ?  Reply With Quote

21. ## Re: VCE Maths questions help Originally Posted by boredsatan is the answer 1/sqrt(3) x + 5 ?
That isn't an equation.  Reply With Quote

22. ## Re: VCE Maths questions help Originally Posted by 1729 That isn't an equation.
y = 1/sqrt(3) x + 5 ?  Reply With Quote

23. ## Re: VCE Maths questions help Originally Posted by boredsatan y = 1/sqrt(3) x + 5 ?
Correct!  Reply With Quote

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?  Reply With Quote

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  Reply With Quote