VCE Maths questions help (2 Viewers)

boredsatan · Apr 23, 2017

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

InteGrand · Apr 23, 2017

Re: maths questions help

boredsatan said:
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.

boredsatan · Apr 23, 2017

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

disturb_equilibrium · Apr 23, 2017

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

the y axis is x=0!!!

boredsatan · Apr 23, 2017

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disturb_equilibrium said:
the y axis is x=0!!!

Oh, right, I get it now, was just a bit confused before.

boredsatan · Apr 24, 2017

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

Green Yoda · Apr 24, 2017

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boredsatan said:
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.

boredsatan · Apr 24, 2017

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Rathin said:
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?

InteGrand · Apr 24, 2017

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boredsatan said:
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!

boredsatan · Apr 24, 2017

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InteGrand said:
Correct!

Thanks

boredsatan · Apr 25, 2017

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

Drongoski · Apr 25, 2017

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boredsatan said:
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.

boredsatan · Apr 29, 2017

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

InteGrand · Apr 29, 2017

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boredsatan said:
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.

boredsatan · Apr 29, 2017

Re: maths questions help

InteGrand said:
$\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

InteGrand · Apr 29, 2017

Re: maths questions help

boredsatan said:
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.$

boredsatan · Apr 29, 2017

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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.

kawaiipotato · Apr 29, 2017

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$\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 $x=\frac{3}{2}$.$

boredsatan · Apr 30, 2017

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

boredsatan · Apr 30, 2017

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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.

VCE Maths questions help (2 Viewers)

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