Curve Sketching (1 Viewer)

[x.Exhaust.x]

Okay, I've got a few more questions to ask .

If f(x)=-x^2+6x-4, sketch:

1. y=|f(x)|
2. y=f(-x)
3. y=1/f(x)
4. y^2=f(x)

Would anyone have tips on approaching locus questions (e.g. shading the area). Any tricks? Lol.

Oh, and for example:

i^2009. Would you have to divide by a common multiple of 4, and then figure out i?.

i^4(502)+1

= i^1

=i

= sqrt i

 

[lyounamu]

Okay, I've got a few more questions to ask .

If f(x)=-x^2+6x-4, sketch:

1. y=|f(x)|
2. y=f(-x)
3. y=1/f(x)
4. y^2=f(x)

Would anyone have tips on approaching locus questions (e.g. shading the area). Any tricks? Lol.

Oh, and for example:

i^2009. Would you have to divide by a common multiple of 4, and then figure out i?.

i^4(502)+1

= i^1

=i

= sqrt i

um..

close...but you got the last line of working incorrectly. It's just i or sqrt (-1)

for the graph parts,

i) use the quadratic formula to factorise the equation.

f(x) = -x^2 + 6x -4 = -(x^2-6x+4)

x = (6 +- sqrt (36- 4 . 1 . 4))/2 = (6 +- sqrt (20))/2 = 3 +- sqrt(5)

so f(x) = -(x-(3+sqrt(5))(x-(3-sqrt(5)
for the absolute value one: test the points above 3+sqrt(5), below 3-sqrt(5), between those values and draw the graph according to that.

so it will be like:

f(x) = -(x-(3+sqrt(5))(x-(3-sqrt(5)) if x > 3+sqrt(5)
f(x) = (x-(3+sqrt(5))(x-(3-sqrt(5)) if 3+sqrt(5)>x>3-sqrt(5)
f(x) = -(x-(3+sqrt(5))(x-(3-sqrt(5)) if x< 3-sqrt(5)

draw those three separate graphs within those domains.

ii) well, that graph is simple enough. just sub -x into x in place for f(x) and work it out using quadratic formula to find the points where it intersects x axis.

iii) I got you the equation. With that you will be able to find vertical asymptotes which are x = 3-sqrt(5) and x=3+sqrt(5) then horizontal asymptote can be also found using x->infinity or other method. Find derivative if you have to (to find stationary points). Box method also helps.

iv) I am thinking that this one will look like two lines heading to opposite direction (looks like log graph or curved line). Just say y = sqrt (f(x)) then work out the asymptotes or the points where it "stops"

 

[Trebla]

This should be in the Extension 2 forum...and btw i is NOT equal to √i.

 

[x.Exhaust.x]

This should be in the Extension 2 forum...and btw i is NOT equal to √i.

LOL √-1 I mean Unconscious error...

Thanks Namu.

 

[cyl123]

for the curve sketching components, once you know how to draw y=f(x), it should be easy to draw the rest
i) |f(x)| : just reflect all parts where f(x) < 0 about the x axis, while leaving the rest of the graph as it is. it uses the principle |f(x)| = f(x) if f(x)>=0, -f(x) if f(x)<0

ii) reflect whole graph y = f(x) about y axis

iii) x intercepts of f(x) ----> vertical asymptotes of y=1/f(x)
if as x---> infinity (test for both + and -) and f(x) ----> g(x), where g(x) is a defined function which the y value approaches (can be function or just a number), then 1/f(x) ---> 1/g(x) (Obvious enough)
Maxima turning pts ---> minima turning pts. (and vice versa)

iv) ignore all x values of y=f(x) where y<0. Obviously x intercepts of f(x) are same as intercepts of y^2 = f(x). the curves are less steep than y=f(x), and the graph is symmetrical about x axis (Obviously as eg. when subbing in y=-1, you're going to get the same result as subbing in y = 1)

 

[youngminii]

For 4. Make sure that all the turning points of f(x) are now cusps. (this will mean you have an inflexion point)
I lost a mark for this in my exam. God damn

 

[pyrodude1031]

See attached sketch.
I used matlab to sketch it.
effectively,

abs , take everything negative and draw it as a mirror reflection onto above the xaxis

f(-x) , draw f(x) 's reflection aboout the yaxis

1/ x intercepts -> vert asymp, analyse positive/negativeness around the intercepts. and determine the horizontal asymp

y^2 , draw the square root of the f(x) and also draw on the same graph its reflection about the xaxis. Because of square root, f(x) whcih is negative is ignored.

Killara High 2005
Internal Dux for MX1 MX2 Physics with HSC marks 98 95 93
B.E. (biomedical) B.Sc. (Advanced Maths) USYD

 

[youngminii]

See attached sketch.
I used matlab to sketch it.
effectively,

abs , take everything negative and draw it as a mirror reflection onto above the xaxis

f(-x) , draw f(x) 's reflection aboout the yaxis

1/ x intercepts -> vert asymp, analyse positive/negativeness around the intercepts. and determine the horizontal asymp

y^2 , draw the square root of the f(x) and also draw on the same graph its reflection about the xaxis. Because of square root, f(x) whcih is negative is ignored.

Killara High 2005
Internal Dux for MX1 MX2 Physics with HSC marks 98 95 93
B.E. (biomedical) B.Sc. (Advanced Maths) USYD

The scale of those axis are way too.. I can't think of the word. Bad?

 

[pyrodude1031]

The scale of those axis are way too.. I can't think of the word. Bad?

Ok.. I must say that I am not giving you a beauty contest. Just a guide as to the shape of the curve and where critical things are.

I simply chucked in a vector of values and the corresponding y vector and plotted them.

I could have changed the axes... But no big deal. If you want, I can post the script for the program and you can download the MATLAB program and do it yourself. Change axes, label them, have colours, have squares as plotting points... many things to change the LOOK of it.

But, what the marker counts is the shape of it. In 4Unit exams, if you do do 4Unit, axes doesnt count towards a great part of the marks.

That does sound bad. But look, not many pplz would spend that time to do a plot, and post an image. I cant see any replies above gave you an image. Just a bunch of text, from which you derive the picture. I think I have been kind enough. I was gonna just tell you in writing which curve is which, I actually went into the trouble of labelling the curves.