# 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

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.

3. ## Re: maths questions help

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

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!!!

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.

6. ## Re: maths questions help

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

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.

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?

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!

10. ## Re: maths questions help

Originally Posted by InteGrand
Correct!
Thanks

11. ## Re: maths questions help

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

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.

13. ## Re: maths questions help

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

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
$\noindent Set the discriminant \Delta := (5m-1)^{2} -4(2m-3)(3m-2) to be greater than 0 and solve for m. In other words, solve the quadratic inequation (5m-1)^{2} -4(2m-3)(3m-2) > 0 for m.$

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

15. ## Re: maths questions help

Originally Posted by InteGrand
$\noindent Set the discriminant \Delta := (5m-1)^{2} -4(2m-3)(3m-2) to be greater than 0 and solve for m. In other words, solve the quadratic inequation (5m-1)^{2} -4(2m-3)(3m-2) > 0 for m.$

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

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
$\noindent Have you learnt how to solve quadratic inequations like this? If not, you should probably learn that first, since you need to know how to do it for this question. Basically you find the roots of the quadratic m^2 + 42m - 23, sketch the parabola, and deduce from there for which values of m the quadratic is positive.$

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.

18. ## Re: maths questions help

$\noindent a. If x is the side-length of the square, then the square has perimeter P_S = 4x.Then the remaining length for the wire used for the rectangle is 12 - 4x. Let a,2a be the dimensions of the rectangle, so the perimeter of the rectangle P_R = 6a. Hence, 6a = 12-4x \implies a = \frac{12-4x}{6}. The dimensions of the rectangle are \frac{12-4x}{6} and \frac{12-4x}{3}.$

$\noindent b. A = x^2 + \frac{12-4x}{6}\frac{12-4x}{3} = x^2 + \frac{(12-4x)^2}{18}.$

$\noindent c. \frac{dA}{dx} = 2x - \frac{8(12-4x)}{18} \\ Let \frac{dA}{dx} = 0 for stationary points \\ 2x - \frac{8(12-4x)}{18} = 0 \implies x = \frac{3}{2}.$

$\noindent You can show this is a minimum by taking the second derivative and showing the sign is positive for x=\frac{3}{2}.$

19. ## Re: maths questions help

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

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.

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).

22. ## Re: maths questions help

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

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 .

24. ## 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.

25. ## 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.
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).
.

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