VCE Methods Question Thread!

[Zues]

Well consider the transformations from the normal
to 

There is a translation by 2 units in the positive directin parallel to the y-axis (up), and there is also a translation of 2 units in the negative direction parallel to the x-axis (left), so the end point is [-2, 2]. With no reflection in the x-axis, the graph normally shoots out to the right, intersecting the y-axis at (0, 2+)

So, to answer your question. How can there be a x-value if the graph actually starts at (-2,2)?

Hope it helped

yes i already understood that there will be no x intercepts.

But mathematically i want to solve that, it might be a knowledge gap which i would like to fix.

[cosine]

yes i already understood that there will be no x intercepts.

But mathematically i want to solve that, it might be a knowledge gap which i would like to fix.

Theres nothing you are missing out on. Its just that when y=0, x= BUT, the endpoint of square root functions is always the start of the domain.

So, if the endpoint is (-2,2) that means the range is [2, infinity). If the graph's range starts at positive 2, there is no x-intercept.

x= only when y=0, but in this specific graph, the lowest y-value is 2, not 0. Hence why there isnt an x-value.

[Zues]

alright then that makes sense. I was just expecting a non existant number on the Re(z) axis.

btw, can someone show me how to graph  y=2−|x^2 −2|

[cosine]

alright then that makes sense. I was just expecting a non existant number on the Re(z) axis.

btw, can someone show me how to graph  y=2−|x^2 −2|

Whenever I sketch absolute graphs I follow these steps:

Step 1:
Rearrange the function (if its hasnt already been) to make sure its in the order

rearranged is

Step 2: Draw the graph that is inside the |....|, that is, draw the function inside the modulus and disregard everything outside of it. Graphing should yield a parabola that intercepts the y-axis at (0,2) and the x-axis at -,+

Step 3: Consider the outside terms now. We have a - sign in front of the modulus sign, so this means everything that is in the positive x-axis (that is, above the x-axis) should be reflected down.

Step 4: Consider any other transformations. Now, we have that annoying +2 at the end, acting as the 'k' of our general equation . So basically, look at your graph and move the whole thing 2 units up.

Thats it. Hope it helped

[Zues]

the x intercepts now change to 2 and -2. why is that

[cosine]

the x intercepts now change to 2 and -2. why is that

Well because the x-intercepts of +,- were of the equation .  But now that we have moved the equation 2 units up and considered the everything outside the mod signs, the new equation becomes Now solve for x by allowing y=0.

[Zues]

and finally with : Step 3: Consider the outside terms now. We have a - sign in front of the modulus sign, so this means everything that is in the positive x-axis (that is, above the x-axis) should be reflected down.

would we not just sketch the graph again but the whole thing reflected in the x axis? i.e graph -x^2-2 ?

[cosine]

and finally with : Step 3: Consider the outside terms now. We have a - sign in front of the modulus sign, so this means everything that is in the positive x-axis (that is, above the x-axis) should be reflected down.

would we not just sketch the graph again but the whole thing reflected in the x axis? i.e graph -x^2-2 ?

Yes..

[Zues]

yeah why is that?

[knightrider]

how is the = i dont understand

[Zealous]

how is the = i dont understand

Take out common factors:

[SE_JM]

Can someone tell me what transformation has occurred to this graph:
x=8 (1-e^-0.2t) (i'm sketching the graph of x against t)

I know there is
-reflection in the x-axis
-reflection in the t-axis
followed by
-a vertical translation of 1 unit in the positive direction of  x-axis
- dilation factor of 5 from x-axis (i'm not sure if i'm right on this one)

But what about the 8 in front? is that a dilation factor?

[cosine]

Can someone tell me what transformation has occurred to this graph:
x=8 (1-e^-0.2t) (i'm sketching the graph of x against t)

I know there is
-reflection in the x-axis
-reflection in the t-axis
followed by
-a vertical translation of 1 unit in the positive direction of  x-axis
- dilation factor of 5 from x-axis (i'm not sure if i'm right on this one)

But what about the 8 in front? is that a dilation factor?

Transformations: DRT (Dilations, Reflections, Translations)

- Dilation by a factor of 8 parallel to the y-axis
- Dilation by a factor 0.2 () parallel to the x-axis

- Reflection in both the x-axis and y-axis
- Translation 8 units in the positive direction parallel to the y-axis

[SE_JM]

Transformations: DRT (Dilations, Reflections, Translations)

- Dilation by a factor of 8 parallel to the y-axis
- Dilation by a factor 0.2 () parallel to the x-axis

- Reflection in both the x-axis and y-axis
- Translation 8 units in the positive direction parallel to the y-axis

thank you

[brightsky]

Transformations: DRT (Dilations, Reflections, Translations)

- Dilation by a factor of 8 parallel to the y-axis
- Dilation by a factor 0.2 () parallel to the x-axis

- Reflection in both the x-axis and y-axis
- Translation 8 units in the positive direction parallel to the y-axis

Be careful with the dilation factors. If you want to turn the exponent from t to 0.2 t, you need to dilate by a factor of 5 from the x-axis, rather than by a factor of 0.2. Also, be careful with how you word your transformations. Although there is nothing technically wrong with the phrase "dilate by a factor of k parallel to the t-axis", it is preferable to write "dilate by a factor of k from the x-axis". As a rule of thumb, we dilate FROM an axis, reflect IN an axis, and translate IN THE POSITIVE/NEGATIVE DIRECTION of an axis. My recommendation would be to write:

1. Dilate by a factor of 8 from the t-axis (which in this case serves as the horizontal axis rather than x). x = e^(t) now becomes x = 8e^(t).
2. Dilate by a factor of 5 from the x-axis (which in this case serves as the vertical axis rather than y). x = 8e^(t) now becomes x = 8e^(t/5) = 8e^(0.2t).
3. Reflect in the t-axis. x = 8e^(0.2t) now becomes x = -8e^(0.2t).
4. Reflect in the x-axis. x = -8e^(0.2t) now becomes x = -8e^(-0.2t).
5. Translate 8 units in the positive direction of the x-axis. x = -8e^(-0.2t) now becomes x = 8-8e^(-0.2t) = 8(1-e^(-0.2t)).

Note that we are not obliged to follow DRT. Another sequence of transformations that converts x = e^(t) to x = 8(1-e^(-0.2t)) is:

1. Dilate by a factor of 5 from the x-axis. x = e^(t) becomes x = e^(t/5) = e^(0.2t).
2. Reflect in the x-axis. x = e^(0.2t) becomes x = e^(-0.2t).
3. Reflect in the t-axis. x = e^(-0.2t) becomes x = -e^(-0.2t).
4. Translate 1 unit in the positive direction of the x-axis. x = -e^(-0.2t) becomes x = 1-e^(-0.2t).
5. Dilate by a factor of 8 from the t-axis. x = 1-e^(-0.2t) becomes x = 8(1-e^(-0.2t)).