VCE Methods Question Thread!

[Anchy]

Let f(x)=x^2 and g(x) = -2x(x-2)+1

State in correct order the transformations necessary to obtain the graph of g(x) from the graph of f(x)

I can't seem to get the right answer! Any help would be appreciated. Thanks

Btw apologize for not using LaTex, don't know how to use it.

[brightsky]

hint: complete the square.

g(x) = -2x^2 + 4x + 1 = -2(x^2 - 2x + 1 - 3/2) = -2[(x-1)^2 - 3/2] = -2(x-1)^2 + 3

therefore:
1. reflect in x-axis
2. dilate by a factor of 2 from the x-axis
3. translate 1 unit in the positive direction of the x-axis
4. translate 3 units in the positive direction of the y-axis

[Anchy]

hint: complete the square.

g(x) = -2x^2 + 4x + 1 = -2(x^2 - 2x + 1 - 3/2) = -2[(x-1)^2 - 3/2] = -2(x-1)^2 + 3

therefore:
1. reflect in x-axis
2. dilate by a factor of 2 from the x-axis
3. translate 1 unit in the positive direction of the x-axis
4. translate 3 units in the positive direction of the y-axis

Oh damn! Thanks heaps for that

[ETTH96]

Hey guys, theres this question, it gives me: f(x)=e^2x  -1
It asks me to find the inverse, I got:
f(x)=loge(x+1) /2
Is this right??

Then it asks me to:
Find f(-f^-1 (2x)) in the form ax/bx+c where a b and c are real constants

How do I do this? Do I sub (2x) into where x is in the inverse function..? Then what..   

[brightsky]

f(x) = e^(2x) - 1
let y = f(x). swap x and y:
x = e^(2y) -1
x+1 = e^(2y)
y = 1/2*ln(x+1)

therefore f^(-1)(x) = 1/2*ln(x+1). your answer looks fine. also note that you can also write f^(-1)(x) = ln(sqrt(x+1)) due to log laws.

f(-f^(-1)(2x))
= e^(2(-1/2 ln(2x+1)) - 1
= e^(ln(1/(2x+1)) - 1
= 1/(2x+1) - 1
= (-2x)/(2x+1)

where a = -2, b = 2, c = 1.

[ETTH96]

Hey brightsky, how did you get from:

f(-f^(-1)(2x))
= e^(2(-1/2 ln(2x+1)) - 1   <-- this
= e^(ln(1/(2x+1)) - 1 <-- to this?

Why does it become 1/(2x+1)? how come the 2x+1 is at the bottom of the fraction I'm a tad confused

[b^3]

.

You end up with a negative out the front, which you can then bring inside the log as a power, which then brings the down as the denominator.

[ETTH96]

.

You end up with a negative out the front, which you can then bring inside the log as a power, which then brings the down as the denominator.

Ohhhhhhhhhhhh I did not know that! Thanks a heap you two   xo

[ETTH96]

How would I solve the intersecting points of these two graphs:

-sqrt(x+8)  +2
and it's inverse,
x^2 - 4x - 4


I let it equal each other but it's getting all messy when I expand it and everything, could someone please tip me off?

[b^3]

The two graphs will intersect somewhere along the line (well there are exceptions but they don't normally come up, so do a quick sketch to make sure it's not one of those cases). This means we can solve either or to get the same solution, given that we make sure we apply the domain restrictions from both functions. So you'd solve over the domain that both the original and inverse are valid for.

[Jason12]

Find lim h->0 (4x^2h^2 +xh +h)/h

[b^3]

Take a factor of out, then you can cancel a set of 's out. Then we want to find the limit of what we have left, that is what happens as we make really small. Since we have a polynomial we can just substitute the value of we have into the expression we have left.

Spoiler

[ETTH96]

Thanks, b^3

How would I find the inverse function for this: f(x)=x^2-6x+5

This is what I've done so far:

x=y^2-6y+5

What do I do from there to solve for y?
x-5=y^2-6y

Then what?! :S

[brightsky]

hint: complete the square.

[ETTH96]

hint: complete the square.

x=y^2-6y+9=9+5
x=(y-3)^2-4
x+4=(y-3)^2
(+-) sqrt(x+4) = y-3

y= 3+ (+-) sqrt (x+4)

THANKS HEAPS BRIGHTSKY! I feel so silly for not thinking of that haha