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April 20, 2024, 06:31:42 am

Author Topic: VCE Methods Question Thread!  (Read 4819046 times)  Share 

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Peas

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Re: VCE Methods Question Thread!
« Reply #17715 on: February 24, 2019, 03:47:48 pm »
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Hi,

Is there a difference between a 'comma' (,) and writing 'or'? For example, the difference between x = 3,4 and x = 3 or 4.

Thanks :)

guac

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Re: VCE Methods Question Thread!
« Reply #17716 on: February 24, 2019, 03:55:05 pm »
+1
Hi,

Is there a difference between a 'comma' (,) and writing 'or'? For example, the difference between x = 3,4 and x = 3 or 4.

Thanks :)

Probably not really but the safest bet is to write 'x = 3 or x = 4'. You just want to make sure it's really clear to the examiner what you're writing.
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aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17717 on: February 24, 2019, 08:30:53 pm »
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Hi! Are all of the \(-\left(x+y\right)^2\ \left(x-y\right)^2\) like equations that have only a square of like one of these (x-3) all only give the turning point? (instead of the x-intercepts because both of them give the same x intercept, like \(-\left(x-1\right)^2)\
If an equation gives me the two same x intercepts, does that mean that the turning point must lie on the x-axis?

AlphaZero

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Re: VCE Methods Question Thread!
« Reply #17718 on: February 24, 2019, 10:02:29 pm »
+1
Hi! Are all of the \(-\left(x+y\right)^2\ \left(x-y\right)^2\) like equations that have only a square of like one of these (x-3) all only give the turning point? (instead of the x-intercepts because both of them give the same x intercept, like \(-\left(x-1\right)^2)\
If an equation gives me the two same x intercepts, does that mean that the turning point must lie on the x-axis?

I think I know what you're trying to ask, but I'll need you to clarify.

Are you essentially asking whether the graph of something of the form  \(y=(x-a)^2(x-b)(x-c)\)  will always have a turning point at  \((a,0)\)  and will always cut the \(x\)-axis at  \((b,0)\)  and  \((c,0)\)? If so, the answer is yes.

To answer your second question: yes. If a polynomial \(P(x)\) has a double root at say \(x=k\), the graph of  \(y=P(x)\)  will have a turning point at \((k,0)\).

In fact, we can generalise this for roots of higher multiplicity.

Suppose that for a polynomial \(P(x)\),  \(\alpha\) is a root with even multiplicity and \(\beta\) is a root with odd multiplicity of at least 3.

That is,  \(P(x)=(x-\alpha)^{2k}(x-\beta)^{2k+1}S(x)\),  where \(k\in\mathbb{N}\) and \(S(x)\) is another polynomial.

Then,  the graph of  \(y=P(x)\)  will have a turning point at  \((\alpha,0)\),  and a stationary point of inflection at  \((\beta,0)\).

Clearly, as mentioned,  \(y=P(x)\)  will cut the \(x\)-axis at single roots.
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aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17719 on: February 26, 2019, 08:15:30 pm »
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I think I know what you're trying to ask, but I'll need you to clarify.

Are you essentially asking whether the graph of something of the form  \(y=(x-a)^2(x-b)(x-c)\)  will always have a turning point at  \((a,0)\)  and will always cut the \(x\)-axis at  \((b,0)\)  and  \((c,0)\)? If so, the answer is yes.

To answer your second question: yes. If a polynomial \(P(x)\) has a double root at say \(x=k\), the graph of  \(y=P(x)\)  will have a turning point at \((k,0)\).

In fact, we can generalise this for roots of higher multiplicity.

Suppose that for a polynomial \(P(x)\),  \(\alpha\) is a root with even multiplicity and \(\beta\) is a root with odd multiplicity of at least 3.

That is,  \(P(x)=(x-\alpha)^{2k}(x-\beta)^{2k+1}S(x)\),  where \(k\in\mathbb{N}\) and \(S(x)\) is another polynomial.

Then,  the graph of  \(y=P(x)\)  will have a turning point at  \((\alpha,0)\),  and a stationary point of inflection at  \((\beta,0)\).

Clearly, as mentioned,  \(y=P(x)\)  will cut the \(x\)-axis at single roots.

ok thanks a lot!!!

aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17720 on: February 27, 2019, 08:50:29 pm »
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Do parabolas of a same family (like \(y=ax^2+bx+12\) ), have the same discriminant for all of them?
Like if it asks a question (giving the equation of like above but without numbers:
"For a parabola in this family with its turning point on the x-axis find a in terms of b"
Am I allowed to use the discriminant that I worked out from the original equation?

Thanks (want to clear this up)

Sine

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Re: VCE Methods Question Thread!
« Reply #17721 on: February 27, 2019, 08:57:18 pm »
+1
Do parabolas of a same family (like \(y=ax^2+bx+12\) ), have the same discriminant for all of them?
think of it logically
The discriminant = b^2 - 4ac
in this case we know c = 12
So discriminant = b^2 - 48a

Will this now be the same for all parabolas in the family?

Like if it asks a question (giving the equation of like above but without numbers:
"For a parabola in this family with its turning point on the x-axis find a in terms of b"
Am I allowed to use the discriminant that I worked out from the original equation?

Thanks (want to clear this up)
From your first line I assume you mean y = ax^2 + bx +c
and your second line - when the turning point of a parabola is on the x-axis it means the parabola has 1 real solution hence discriminant = 0

aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17722 on: February 27, 2019, 09:13:31 pm »
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think of it logically
The discriminant = b^2 - 4ac
in this case we know c = 12
So discriminant = b^2 - 48a

Will this now be the same for all parabolas in the family?
From your first line I assume you mean y = ax^2 + bx +c
and your second line - when the turning point of a parabola is on the x-axis it means the parabola has 1 real solution hence discriminant = 0
I see!
So the discriminant will be the same?

Sine

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Re: VCE Methods Question Thread!
« Reply #17723 on: February 27, 2019, 09:17:03 pm »
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I see!
So the discriminant will be the same?
try subbing in random numbers to try to disprove it

aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17724 on: February 27, 2019, 09:27:55 pm »
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try subbing in random numbers to try to disprove it
Uhh.. this is  making it more confusing for me -_- . I'm moreover looking for a yes or no answer ..

I'm assuming it is correct but from the answers they rearranged a in terms of b from the discriminant.
I hope someone can confirm if this is correct.

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Re: VCE Methods Question Thread!
« Reply #17725 on: February 27, 2019, 09:38:32 pm »
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What do you mean by the same family? I can define a family of parabolas in so many ways:
Equivalent up to translation
Equivalent up to rotation about the origin
Equivalent up to any linear transformation (dilations, reflections, rotations)
Equivalent up to any linear transformation and translation (an affine transformation)
Same turning point
Same x intercepts

What's the family defined as?
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aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17726 on: February 27, 2019, 09:53:52 pm »
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What do you mean by the same family? I can define a family of parabolas in so many ways:
Equivalent up to translation
Equivalent up to rotation about the origin
Equivalent up to any linear transformation (dilations, reflections, rotations)
Equivalent up to any linear transformation and translation (an affine transformation)
Same turning point
Same x intercepts

What's the family defined as?
A family as in having the same form, a same constant (in this case)
\(y=ax^2+bx+10\)
Like having different a and b but the same constant.
Will the discriminant (in this case \(b^2+40a\) ), work with all the other parabolas in this family (same constant of 10 same form y=ax^2+bx+c, but just different number like \(y=12x^2+53x+10\) and \(y=2x^2+3x+10\) ?
edit = when we are not being given the a and b in from y=ax^2+bx+10

AlphaZero

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Re: VCE Methods Question Thread!
« Reply #17727 on: February 28, 2019, 06:47:37 pm »
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A family as in having the same form, a same constant (in this case)
\(y=ax^2+bx+10\)
Like having different a and b but the same constant.
Will the discriminant (in this case \(b^2+40a\) ), work with all the other parabolas in this family (same constant of 10 same form y=ax^2+bx+c, but just different number like \(y=12x^2+53x+10\) and \(y=2x^2+3x+10\) ?
edit = when we are not being given the a and b in from y=ax^2+bx+10

I'm too sure what you mean by "work with all the other parabolas in this family".

Consider the functions  \(f(x)=ax^2+bx+10\),  where \(a,b\in\mathbb{R}\) and \(a\neq 0\).

The discriminant is given by  \(\Delta=b^2-40a\).

So, if we were given say  \(a=2\)  and  \(b=3\),  (that is,  \(f(x)=2x^2+3x+10\)),  then  \(\Delta=3^2-40\!\times\! 2=-71\).
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Re: VCE Methods Question Thread!
« Reply #17728 on: February 28, 2019, 09:09:58 pm »
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would anyone be able to help with this q please!

Given that x^3 − 2x^2 + 5 = ax(x−1)^2 + b(x−1) + c for all real numbers x, find the values of a, b and c.
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Gogurt

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Re: VCE Methods Question Thread!
« Reply #17729 on: February 28, 2019, 09:15:04 pm »
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Not sure how to approach this question: "If the solutions of x^2 + bx + 15 = 0 are integers less than 10 but greater than –10, what are the possible values of b?". I've tried a few thing but no luck.
 Any thoughts?