# Thread: VCE Maths questions help

1. ## VCE Maths questions help

1. How to find the turning point and type of turning point in the equation y = (2/3)x^4 + 1/3
2. how to find equation of axis of symmetry of y = (2/3)x^4 + 1/3  Reply With Quote

2. ## Re: maths questions help Originally Posted by boredsatan 1. How to find the turning point and type of turning point in the equation y = (2/3)x^4 + 1/3
2. how to find equation of axis of symmetry of y = (2/3)x^4 + 1/3
1) Turning point at 0, local minimum.

2) Axis of symmetry is the y-axis.

Method used: inspection.  Reply With Quote

3. ## Re: maths questions help

2) My book said that the axis of symmetry is x = 0. Maybe my book's wrong?  Reply With Quote

4. ## Re: maths questions help Originally Posted by boredsatan 2) My book said that the axis of symmetry is x = 0. Maybe my book's wrong?

the y axis is x=0!!!  Reply With Quote

5. ## Re: maths questions help Originally Posted by disturb_equilibrium the y axis is x=0!!!
Oh, right, I get it now, was just a bit confused before.  Reply With Quote

6. ## Re: maths questions help

What's the difference between the turning point and the stationary point of inflection?  Reply With Quote

7. ## Re: maths questions help Originally Posted by boredsatan What's the difference between the turning point and the stationary point of inflection?
A point where a function changes from an increasing to a decreasing function or visa-versa is known as a turning point (i.e. gradient =0) However with turning points the concavity remains the same. In a stationary point of inflexion the gradient is 0 but the concavity changes, thus not changing from an increasing to a decreasing function or visa-versa.  Reply With Quote

8. ## Re: maths questions help Originally Posted by Rathin A point where a function changes from an increasing to a decreasing function or visa-versa is known as a turning point (i.e. gradient =0) However with turning points the concavity remains the same. In a stationary point of inflexion the gradient is 0 but the concavity changes, thus not changing from an increasing to a decreasing function or visa-versa.
So for x^2, x^4, x^6, x^8, etc, there's a turning point, and for x^3, x^5, x^7, x^9, etc there's a stationary point of inflection?  Reply With Quote

9. ## Re: maths questions help Originally Posted by boredsatan So for x^2, x^4, x^6, x^8, etc, there's a turning point, and for x^3, x^5, x^7, x^9, etc there's a stationary point of inflection?
Correct!  Reply With Quote

10. ## Re: maths questions help Originally Posted by InteGrand Correct!
Thanks   Reply With Quote

11. ## Re: maths questions help

How hard it to get raw 40 in vce maths methods?  Reply With Quote

12. ## Re: maths questions help Originally Posted by boredsatan How hard it to get raw 40 in vce maths methods?
Very few here are familiar with the VCE. But I find the VCE Maths curriculum modern and progressive.  Reply With Quote

13. ## Re: maths questions help

Find the values of m if (2m-3)x^2 + (5m-1)x + (3m-2) = 0 has 2 solutions  Reply With Quote

14. ## Re: maths questions help Originally Posted by boredsatan Find the values of m if (2m-3)x^2 + (5m-1)x + (3m-2) = 0 has 2 solutions

Remember, if the given quadratic equation has two distinct (real) solutions for x, then its discriminant must be greater than zero.  Reply With Quote

15. ## Re: maths questions help Originally Posted by InteGrand Remember, if the given quadratic equation has two distinct (real) solutions for x, then its discriminant must be greater than zero.
I ended up getting 25m^2 - 10m + 1 - 4(6m^2 - 13m + 6)
= 25m^2 - 10m + 1 - 24m^2 + 52m - 24
= m^2 + 42m - 23 > 0  Reply With Quote

16. ## Re: maths questions help Originally Posted by boredsatan I ended up getting 25m^2 - 10m + 1 - 4(6m^2 - 13m + 6)
= 25m^2 - 10m + 1 - 24m^2 + 52m - 24
= m^2 + 42m - 23 > 0  Reply With Quote

17. ## Re: maths questions help

A piece of wire 12 cm long is cut into two pieces. One piece is used to form a square shape and the other a rectangle shape in which the length is twice the width.
a. If x cm is the side length of the square, write down the dimensions of the rectangle in terms of x
b. formulate a rule for A, the combined area of the square and rectangle in cm^2, in terms of x.
c. determine the lengths of the two pieces if the sum of the areas is to be a minimum.  Reply With Quote

18. ## Re: maths questions help  Reply With Quote

19. ## Re: maths questions help

Is it normal to find maths hard even after tution? I'm in this scenario  Reply With Quote

20. ## Re: maths questions help

The graph of y = x^4 - 2x - 12 has 2 x-intercepts
a. construct a table of values for this polynomial rule for x = -3,-2,-1,0,1,2,3
b. Hence state an exact solution to the equation x^4 - 2x - 12 = 0
c. State an interval within which the other root of the equation lies and use the methods of bisection to obtain an estimate of this root correct to 1 decimal place

I get how to do part a and b, but i'm finding part c extremely challenging and confusing.  Reply With Quote

21. ## Re: maths questions help Originally Posted by boredsatan The graph of y = x^4 - 2x - 12 has 2 x-intercepts
a. construct a table of values for this polynomial rule for x = -3,-2,-1,0,1,2,3
b. Hence state an exact solution to the equation x^4 - 2x - 12 = 0
c. State an interval within which the other root of the equation lies and use the methods of bisection to obtain an estimate of this root correct to 1 decimal place

I get how to do part a and b, but i'm finding part c extremely challenging and confusing.
If you know a place where the polynomial is positive and another place where it is negative (and the root you already found does not lie between these two places), then the other root lies between these two numbers. You can then use the bisection method (check your textbook if you haven't learnt it yet).  Reply With Quote

22. ## Re: maths questions help

I read my textbook but I still don't understand the bisection methods.  Reply With Quote

23. ## Re: maths questions help Originally Posted by boredsatan I read my textbook but I still don't understand the bisection methods.
Are there any examples in the textbook using bisection method? See this page if not (there's an example in it): https://en.wikipedia.org/wiki/Bisection_method .  Reply With Quote

24. ## Re: maths questions help

Can someone please help?
The graph of y = x^4 - 2x - 12 has 2 x-intercepts
a. construct a table of values for this polynomial rule for x = -3,-2,-1,0,1,2,3
b. Hence state an exact solution to the equation x^4 - 2x - 12 = 0
c. State an interval within which the other root of the equation lies and use the methods of bisection to obtain an estimate of this root correct to 1 decimal place

I get how to do part a and b, but i'm finding part c extremely challenging and confusing.  Reply With Quote

25. ## Re: maths questions help Originally Posted by boredsatan Can someone please help?
The graph of y = x^4 - 2x - 12 has 2 x-intercepts
a. construct a table of values for this polynomial rule for x = -3,-2,-1,0,1,2,3
b. Hence state an exact solution to the equation x^4 - 2x - 12 = 0
c. State an interval within which the other root of the equation lies and use the methods of bisection to obtain an estimate of this root correct to 1 decimal place

I get how to do part a and b, but i'm finding part c extremely challenging and confusing. Originally Posted by InteGrand If you know a place where the polynomial is positive and another place where it is negative (and the root you already found does not lie between these two places), then the other root lies between these two numbers. You can then use the bisection method (check your textbook if you haven't learnt it yet).
.  Reply With Quote

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