VCE Methods Question Thread!

[Lear]

I believe so as you basically divided throughout by -1

[RuiAce]

I believe so as you basically divided throughout by -1

Implying that they are not the same, but rather the negative of each other.

[Lear]

Ah yes my bad. For some reason by brain instantly thought about solving it for 0 and automatically decided it didn’t really matter :’)

[snowisawesome]

Implying that they are not the same, but rather the negative of each other.

But they're still the same in terms of factorising?

[RuiAce]

But they're still the same in terms of factorising?

One of them will have an extra minus in front.

[Lear]

If i’m not mistaken they are different as you have a negative sign on the outside of one and not the other. (X-3)(X+1) =\= -(X-3)(X+1)

[zhen]

Basically, when you solve the equations for their x-intercepts, or when you let the equations equal 0 and solve them, they will get the same result. However, this is not because they are the same equation. Rather, it is because when you reflect something in the x-axis, the x-intercepts remain the same. So, y=(x+1)(x+2) is not the same as y=-(x+1)(x+2) as the second graph is the first one relected in the x-axis. You can verify this by subbing in an x value. Let x=1. The first equation gives y=6 and the second equation gives y=-6. If they were the same equation and had the same graph, they should get the same y value for each x value inputted into the equation. So, it is clear that the equations are not the same cause it doesn’t satisy this condition. It’s like how the graph of y=x² is different to the graph of y=-x². So, basically the equations aren’t the same unless they are completely identical.

[snowisawesome]

Thanks guys
Does anyone know how to find the inverse of y = 9x - x^3
so x = 9y-y^3
x = y(9-y^2)
stuck after this step

[VanillaRice]

Thanks guys
Does anyone know how to find the inverse of y = 9x - x^3
so x = 9y-y^3
x = y(9-y^2)
stuck after this step

You can complete the square for y, which will give you a single y value to make as the subject
Alternatively, you could have also completed the square for x in the first place.

Hope this helps

[snowisawesome]

You can complete the square for y, which will give you a single y value to make as the subject
Alternatively, you could have also completed the square for x in the first place.

Hope this helps

Do you know how to complete the square for a cubic function in this case?

[RuiAce]

You can complete the square for y, which will give you a single y value to make as the subject
Alternatively, you could have also completed the square for x in the first place.

Hope this helps

Check the question again; it's a cubic (in fact with two stationary points). If the domain of that function is \(\mathbb{R}\) then it's not one-to-one

[VanillaRice]

Do you know how to complete the square for a cubic function in this case?

Check the question again; it's a cubic (in fact with two stationary points). If the domain of that function is \(\mathbb{R}\) then it's not one-to-one

My bad - I read the question as a quadratic.

Even if there was a domain restriction, I don't think there is a way to do this in VCE methods - may I ask where the question is from?

[snowisawesome]

according to my textbook -√(16-x^2) is a one - to - one function? How's that possible? Shouldn't it be many to one?

[zhen]

according to my textbook -√(16-x^2) is a one - to - one function? How's that possible? Shouldn't it be many to one?

Just wondering, but are there any domain restrictions applied to this function? For example, that would be a one to one function if the domain=[0,4]

[snowisawesome]

Just wondering, but are there any domain restrictions applied to this function? For example, that would be a one to one function if the domain=[0,4]

It specified that the domain of it's inverse was [-4,0]