Author Topic: VCE Methods Question Thread! (Read 5637286 times) Tweet Share

Sanguinne · « **Reply #1965 on:** May 19, 2013, 04:32:03 pm »

where did you get the point [0,200] from?

Homer · « **Reply #1966 on:** May 19, 2013, 04:43:03 pm »

i'm not sure what you mean, but for C(n) the point is (0,0)

Sanguinne · « **Reply #1967 on:** May 19, 2013, 04:45:39 pm »

sorry i didnt mean (0,200) but (0,0)
where did the (0,0) come from

Homer · « **Reply #1968 on:** May 19, 2013, 04:48:20 pm »

Came from Timmeh

Quote from: Timmeh on May 19, 2013, 03:41:31 pm

It costs $0 to make 0 items

jono88 · « **Reply #1969 on:** May 19, 2013, 05:30:12 pm »

Describe in order the transformations required to transform the graph of y=-2/x to y=1/2(x+5) +6

Alwin · « **Reply #1970 on:** May 19, 2013, 05:34:49 pm »

Quote from: jono88 on May 19, 2013, 05:30:12 pm

Describe in order the transformations required to transform the graph of y=-2/x to y=1/2(x+5) +6

Assuming for the second equation you mean: y=1/(2(x+5)) +6,

Reflect in the x-axis
Dilate by a factor of 1/4 from the x axis
Translate -5 units parallel to the x axis
Translate +6 units parallel to the y axis

jono88 · « **Reply #1971 on:** May 19, 2013, 05:46:43 pm »

Cool thanks so much.

jono88 · « **Reply #1972 on:** May 19, 2013, 08:36:04 pm »

f(x)=loge(x) and g(x)=

Does g[f(x)] exist?

BubbleWrapMan · « **Reply #1973 on:** May 19, 2013, 08:42:55 pm »

You mean $|x|$ ? Well, the domain of $g$ is $\mathbb{R}$ , so $g\circ f$ is defined for any $f$ .

Markkiieee · « **Reply #1974 on:** May 19, 2013, 10:40:43 pm »

Yo!

Okay so this is from Essential methods 3\4, 9G, 2a).

"Find the y-coordinate and the gradient at the points of each of the following curves corresponding to the given value of x:"

Question not typable, the question is attached as a picture below.

b^3 · « **Reply #1975 on:** May 19, 2013, 10:47:24 pm »

$y$ value, substitute in the value for $x$ .
$\begin{alignedat}{1}x=1 \\ y & =\left(2+1\right)^{4}\left(1\right)^{2} \\ & =81 \end{alignedat} $
For the gradient apply the product rule with
$\begin{alignedat}{1}u & =\left(2x+1\right)^{4} \\ \frac{du}{dx} & =4\left(2x+1\right)^{3}\times2 \\ v & =x^{2} \\ \frac{dv}{dx} & =2x \end{alignedat} $
$\begin{alignedat}{1}\frac{dy}{dx}= & u\frac{dv}{dx}+v\frac{du}{dx} \\ \frac{dy}{dx} & =\left(2x+1\right)^{4}\times\left(2x\right)+x^{2}\times4\left(2x+1\right)^{3}\times2 \\ & =2x\left(2x+1\right)^{3}\left(2x+1+4x\right) \\ & =2x\left(2x+1\right)^{3}\left(6x+1\right) \\ x=1 \\ \implies\frac{dy}{dx} & =2\left(1\right)\left(2+1\right)^{3}\left(6+1\right) \\ & =2\times27\times7 \\ & =378 \end{alignedat} $

Jeggz · « **Reply #1976 on:** May 19, 2013, 10:51:46 pm »

Quote from: Markkiieee on May 19, 2013, 10:40:43 pm

Yo!

Okay so this is from Essential methods 3\4, 9G, 2a).

"Find the y-coordinate and the gradient at the points of each of the following curves corresponding to the given value of x:"

Question not typable, the question is attached as a picture below.

So let's address finding the y-coordinate of the point first. Substituting $x=1$ into the equation $y=(2x+1)^4 x^2$ gives the value of y to be $81$ .

Secondly, we need to find the gradient at the point $x=1$ . We can differentiate $y=(2x+1)^4 x^2$ using the product rule
$\frac{dy}{dx}$ $=(2x+1)^4 2x + x^2(8(2x+1)^3)$

$= 2x(2x+1)^4 + 8x^2 (2x+1)^3$

We can tidy this expression up by taking out the common factor of $2x(2x+1)^3$
$2x(2x+1)^3 [(2x+1+4x)]$

$=2x(2x+1)^3 [(6x+1)]$

Now that we have the derivative we can substitude in $x=1$ to find the gradient at the point.
Hence $2(3)^3(7) = 378$

Hopefully that answer is correct - apologies for any mistakes in advance! Hope that made sense too

EDIT: Beaten

Markkiieee · « **Reply #1977 on:** May 19, 2013, 10:54:57 pm »

Thank you both b^3 and Jeggz

clıppy · « **Reply #1978 on:** May 20, 2013, 10:54:44 pm »

Quick transformation question:

Describe the transformations of the graph
y = cos(x) to y= -1/4*cos(4pi(x-(1/2)))+1/3

I'm confused as to when I'm doing, after applying the dilations and the reflections when applying the transformation in the x-axis do i need to take the $4\pi$ into account?

Jeggz · « **Reply #1979 on:** May 20, 2013, 11:09:19 pm »

Quote from: noclip on May 20, 2013, 10:54:44 pm

Quick transformation question:

Describe the transformations of the graph
y = cos(x) to y= -1/4*cos(4pi(x-(1/2)))+1/3

I'm confused as to when I'm doing, after applying the dilations and the reflections when applying the transformation in the x-axis do i need to take the $4\pi$ into account?

1. Dilation of factor $\frac{1}{4}$ from the x-axis.

2. Dilation of factor $\frac{1}{4\pi}$ from the y-axis.

3. Reflection in the x-axis

4. Translation of $\frac{1}{2}$ in the positive x-direction.

5. Translation of $\frac{1}{3}$ in the positive y direction

Hope that helps

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