VCE Methods Question Thread!

[Springyboy]

Alright so my school's already begun methods 3/4 for next year so can someone help me with this question

Im looking to find the inverse of f(x)=x^2+2x with a domain of [-1, ∞) on the original function

[Springyboy]

Where is this question from? From inspection it seems very tedious and not really part of the methods course

Im pretty sure it was from maths quest 11 but dont worry i dont think i need to know that for methods 3/4 and ive already moved on to 3/4 anyway

[Sine]

Alright so my school's already begun methods 3/4 for next year so can someone help me with this question

Im looking to find the inverse of f(x)=x^2+2x with a domain of [-1, ∞) on the original function

For inverses we are trying to reflect the graph across the line y=x this is done by swapping x and y

f(x)=x^2+2x
f(x)=y
so y=x^2+2x (swap x and y)
x=y^2+2y
x=(y+1)^2-1
x+1=(y+1)^2
sqrt(x+1)-1=y
y=+or -sqrt(x+1)-1
we take the - graph + graph as the we are using the leftright side of the parabola
so f-1(x)=sqrt(x+1)-1 (inverse)

EDIT: read domain of original function incorrectly - Thanks hamo94

[Orb]

Alright so my school's already begun methods 3/4 for next year so can someone help me with this question

Im looking to find the inverse of f(x)=x^2+2x with a domain of [-1, ∞) on the original function

With reference to sine, I'm pretty sure the negative function is incorrect and that's the one you need to reject.
Range of original function is increasing and becoming more positive

edit: solutions attached

[Springyboy]

Thanks for your help just had to factorise the y side but i got there in the end

[John_]

Hey guys, can you help me with this question?

What value for k would make y = –x tangent to the curve with equation y = kx – 1?
A: k = 1
B: k > 1
C: k < 1/4
D: k = 1/4
E: 4k - 1 < 0

If possible can you show me the necessary working out. Thanks so much guys!

Along with the original question, I think the answers are wrong.

The graph of y=-x is just a straight line. While the graph of y= x^2 - k x + 4 is a parabola.
If a straight line is a tangent to a parabola, it means that they only intersect at one point, and at one point only due to the exponential nature of parabolas. Hence, we equate the two equations and use the discriminant to find the values of k when the graphs only intersect once.

y=-x      y=x^2 + kx + 4
-x = x^2 - kx + 4
0 = x^2 + x - kx + 4
0 = x^2 + (1-k)x + 4

let the discriminant equal 0 to find when x only has a single solution

0 = b^2 - 4ac
in this case a=1 b=(1-k) and c=4
0 = (1-k)^2 - 4(1)(4)
0 = 1 - 2k + k^2 -16
0 = k^2 - 2k -15
0 = (k-5)(k+2)     
k= -2 and 5 

Hope this helped @Bedigursimran

[John_]

Getting a bit stuck on this question:

The letters of the word featuring are randomly rearranged. Find the probability that the letters of the word feat are together, though not necessarily in the order shown.

Would the answer be 3/182?

If I am right, I did it using permutations. I'll elaborate more if you can confirm the answer.

[lzxnl]

Along with the original question, I think the answers are wrong.

The graph of y=-x is just a straight line. While the graph of y= x^2 - k x + 4 is a parabola.
If a straight line is a tangent to a parabola, it means that they only intersect at one point, and at one point only due to the exponential nature of parabolas.

Parabolas are not exponential. They're quadratic. They grow far slower than exponentials and this distinction is very important in other fields of maths. For instance, an exponential grows faster than any polynomial you can possibly think of.

Secondly, your statement that tangent -> intersect at one point is correct, but note that the converse isn't true. The normal to the vertex of the parabola will also only intersect it once.

Just a few things I thought I'd share.

[Jay.C]

Could someone please explain how to do the following question:

Y= 150 when x=0

Find a b c and d from the following equation:

 y=3ax^3 + 2bx^2 + cx + d.

Thanks guys! 

[TheAspiringDoc]

Could someone please explain how to do the following question:

Y= 150 when x=0

Find a b c and d from the following equation:

 y=3ax^3 + 2bx^2 + cx + d.

Thanks guys!

Doesn't make sense to me as:
150=3a0^3+2b0^2+c0+d
From this, we can deduce that d=150 as the other three parts of the RHS just become 0.
But how to find a,b and c, I have no idea..?

[zsteve]

No, there's no solution. You can only deduce value of d, nothing more. Infinitely many curves of the specified form have y(0)=150

[MightyBeh]

Could someone please explain how to do the following question:

Y= 150 when x=0

Find a b c and d from the following equation:

 y=3ax^3 + 2bx^2 + cx + d.

Thanks guys!

Are you sure you're not missing something?  Usually a question like this would give you a value of dy/dx or something. As a rule of thumb, you need one piece of information per unknown, so as far as I'm aware we can only find d.

[Jay.C]

Sorry for the confusion guys, wrote the total wrong question. Could somone help me with how to do this?

Y=ax^2 + bx + c passes through (2,-4) and had a stationary point at (1,-3/2). Find a b and c.

[Sine]

Sorry for the confusion guys, wrote the total wrong question. Could somone help me with how to do this?

Y=ax^2 + bx + c passes through (2,-4) and had a stationary point at (1,-3/2). Find a b and c.

so it seems that you've been given 2 pieces of information for 3 unknowns but their are actually 3.

2 pieces of info are in the form of points the curve passes through (2,-4) & (1,-3/2) but the 3rd piece of info lies in the fact that they have stated that the point (1,-3/2) is also a stationary points meaning that dy/dx given x = 1 is equal to 0

uses that info to formulate 3 separate equations and solve from there.

[@#035;3]

So..
What you're given is...
f(2)= -4, f(1)=-3/2, f'(1)=0 (stationary points occur @ a zero gradient)
So you'll have to solve 3 simultaneous equations..

Let y=f(x)- I just prefer to use f(x)
f(x)=a*x^2+b*x+c
f(2)=-4
->> 4*a+2*b+c=-4  (equation 1)

f(1)=-3/2
->>a+b+c=-3/2 (equation 2)

f'(1)=0 (because Zero gradient @ stationary points)
f'(x)=2*a*x+b
-->> 2*a+b=0 (equation 3)

Now solve..IDK how to solve 3 simultaneous equations so I just used cas..

which gave me...    a=-5/2, b=5, c=-4


Hope that's correct and hopefully I was helpful.