VCE Methods Question Thread!

[brightsky]

Not really, but I'd recommend starting with Functions and Relations, then proceeding onto Calculus, and ending with Probability. The reason is that calculus cannot really be done without knowledge of functions and probability (in particular the section on continuous random variables) cannot really be done without knowledge of calculus.

[knightrider]

Not really, but I'd recommend starting with Functions and Relations, then proceeding onto Calculus, and ending with Probability. The reason is that calculus cannot really be done without knowledge of functions and probability (in particular the section on continuous random variables) cannot really be done without knowledge of calculus.

Thanks brightksy

[knightrider]

How is this graph a many to many relation?

I thought it was a many to one relation as for any x value there is only one y value
and for any y value there is 2 x values

[Sunshine98]

How is this graph a many to many relation?

I thought it was a many to one relation as for any x value there is only one y value
and for any y value there is 2 x values

Its because it has a shaded region , so for each x-value , there are , in fact,  many y values.
Because there is a region , your looking at what falls within the region , framed by the parabola and not the actual parabola  , alone.
I hope that makes sense 

[knightrider]

Its because it has a shaded region , so for each x-value , there are , in fact,  many y values.
Because there is a region , your looking at what falls within the region , framed by the parabola and not the actual parabola  , alone.
I hope that makes sense 

Thanks Sunshine98 but what does exactly the shaded region do to the graph when determining the type of relation

[brightsky]

A relation is defined simply as a set of points on the Cartesian plane. Since a curve can be thought of as consisting of an infinite number of points, any random squiggle that you draw on the Cartesian plane is, by definition, a relation. As you know, we classify relations based on how many x-values pair up with or correspond to how many y-values. A many-to-one relation, for example, is a relation where many x-values pair up with the one y-value. It is important to realise that the relation you have provided is not actually a parabola (which, I concur, is a many-to-one relation). The relation is in fact a shaded region on the Cartesian plane, which so happens to be bound below by a parabola. The points that make up the relation in this case do not arrange themselves into a nice curve. Rather, they congregate together to fill in an entire region. Now, ask: how many x-values correspond to how many y-values? Consider y = 0. How many x-values correspond to y = 0? Infinitely many! Consider x = 0. How many y-values correspond to x = 0? Again, infinitely many! Hence, the relation in this case is a many-to-many relation, rather than a many-to-one relation.

[Zues]

why does -x+5 turn into x-5, i dont understand this, however i understand the plus before it

[nhmn0301]

why does -x+5 turn into x-5, i dont understand this, however i understand the plus before it

so you have - (-x+5), if you take the negative sign out, it - - (x-5) = + (x-5)

[Zues]

a and b please

[AirLandBus]

a and b please

Hahaha, i was just about to the post the same question. Doing the same questions!

[nhmn0301]

a and b please

a) Sub x=1 in the equation, 1 + 3 + k - k - 4 is 0.
b) if x=1 is a solution, you can rewrite the equation as (x-1) (x^2 + bx + (k+4)) =0 since we want x=1 as the only solution, we need to find condition for x^2 + bx + c cannot equal 0. Expand the above equation and then equate coefficient, you see that b=4. So we have x^2 + 4x + (k+4), for this function not to equal 0, determinant should be <0. thus solve the inequality 16 - 4(k+4) <0 and we have k<0 is the condition.

[brightsky]

a) Let P(x) = x^3 + 3x^2+ kx - k - 4. Note that P(1) = 0, so x = 1 is always a solution.
b) P(x) = (x - 1) (x^2 + 4x + k + 4). For x = 1 to be the only solution, the discriminant of the quadratic factor must be less than 0. 4^2 - 4(k+4) = -4k < 0. So k > 0.

[nhmn0301]

a) Let P(x) = x^3 + 3x^2+ kx - k - 4. Note that P(1) = 0, so x = 1 is always a solution.
b) P(x) = (x - 1) (x^2 + 4x + k + 4). For x = 1 to be the only solution, the discriminant of the quadratic factor must be less than 0. 4^2 - 4(k+4) = -4k < 0. So k > 0.

Oops thanks brightsky, I stuffed up the inequality sign!

[AirLandBus]

I missed out on methods today so i didnt learn chapter 1.2. Can anyone just quickly run over by what the picture means. In particular when to use closed and when to use open brackets for interval notation. Thanks.

[keltingmeith]

I missed out on methods today so i didnt learn chapter 1.2. Can anyone just quickly run over by what the picture means. In particular when to use closed and when to use open brackets for interval notation. Thanks.

(Image removed from quote.)

Closed brackets mean you include the point at the end - open brackets mean you don't.

So for your example, they've defined the domain such that you don't include x=-3 or x=3. This means that when you find the range, you'll notice that the lowest point, y=-3, nothing weird is happening - but, the highest point, y=0, occurs at x=-3 and x=3 - which aren't defined in the domain. So, y=0 isn't defined in the range. This means you must use open brackets around the 0, because it's not defined and we're not including it.