Calculus: Curve sketching (1 Viewer)

[Makro]

1. Find the turning points on the curve: y = (x - 4)(x + 2)² and hence sktech the curve.

Does this use the product rule? I tried that but then I got a funny equation, and couldn't get anywhere.

2. Find all stationary points and inflexions on the curve: y = (2x + 1)(x - 2)⁴. Sketch the curve.

Once I understand the 1st question, would this be the same?

3. Find any stationary points on the curve: y= x²/ x - 2. By also considering the domain of the curve, sketch its graph.

Just quite confused with this one, last question in the exercise. Never get them..

Thank you in advance.

 

[lolokay]

yeah just use the product rule for the first two + factorise to find the stat. points
it shouldn't give you anything weird

for 3, you have an asymptote at x=2 (since domain is all real x, x=/=2
differentiate with product/quotient rule to get the stat points + whether they're max/min, and to see if there's inflections

i don't know whether you need to consider it, but there's an asymptote at y=x, as y->x as x->infinity

then just draw in the graph, considering the asymptotes and min/max (it will be a hyperbola)


for 1.


so stat points at x= +-2

 

[tacogym27101990]

Does this use the product rule? I tried that but then I got a funny equation, and couldn't get anywhere.


Once I understand the 1st question, would this be the same?

Just quite confused with this one, last question in the exercise. Never get them..

Thank you in advance.

1. u=x-4 , u'=1 , v=(x+2)², v'=2(x+2)
hence dy/dx = (x+2)²+(x-4).2(x+2)
=(x+2)[(x+2)+2x-8]
=(x+2)(3x-6)
=3(x+2)(x-2)
now for stat. points dy/dx = 0
so x=2, -2
test for their nature
2. exactly the same.
3. This involves the quotient rule
ie. d/dx(u/v)= (vu'-uv')/v²
so if y = x²/(x-2)
u=x², u'=2x, v=x-2, v'=1
so dy/dx= ((x-2).2x-(x²))/(x-2)²
=(2x²-4x-x²)/(x-2)²
=(x²-4x)/(x-2)²
for stat. points dy/dx =0
therefore x²-4x=0
x(x-4)=0
stat points are 0 , 4. determine their nature
realise also that there is a vertical asymptote at x=2. ie. x=/=2
and also an oblique asymptote at y=x

give me some rep points??

 

[tommykins]

1. u=x-4 , u'=1 , v=(x+2)², v'=2(x+2)
hence dy/dx = (x+2)²+(x-4).2(x+2)
=(x+2)[(x+2)+2x-8]
=(x+2)(3x-6)
=3(x+2)(x-2)
now for stat. points dy/dx = 0
so x=2, -2
test for their nature
2. exactly the same.
3. This involves the quotient rule
ie. d/dx(u/v)= (vu'-uv')/v²
so if y = x²/(x-2)
u=x², u'=2x, v=x-2, v'=1
so dy/dx= ((x-2).2x-(x²))/(x-2)²
=(2x²-4x-x²)/(x-2)²
=(x²-4x)/(x-2)²
for stat. points dy/dx =0
therefore x²-4x=0
x(x-4)=0
stat points are 0 , 4. determine their nature
realise also that there is a vertical asymptote at x=2. ie. x=/=2
and also an oblique asymptote at y=x

give me some rep points??

repped you for lulz

 

[tacogym27101990]

repped you for lulz

haha cheers

 

[Makro]

3. When determining their nature, for x=0 I got neg on both sides meaning an inflection, but that's obviously wrong. Anyone shed some light?

Thanks for the answers so far.

 

[tacogym27101990]

3. When determining their nature, for x=0 I got neg on both sides meaning an inflection, but that's obviously wrong. Anyone shed some light?

Thanks for the answers so far.

f'(x)= (x²-4x)/(x-2)²
f'(1)= (1-4)/4
=-3/4
f'(-1)=(1+4)/4
=5/4
therefore maximum

 

[x3.eddayyeeee]

how do you determine if there is oblique asymptoes?

 

[tommykins]

You normally stumble upon them when finding horizontal assymptotes.

Either polynomial division or algebraic manipulation.

 

[x3.eddayyeeee]

You normally stumble upon them when finding horizontal assymptotes.

Either polynomial division or algebraic manipulation.

thanks again.
i understand what to look for with algebraic manipulation
but what are general results for oblique assymptoes for polynomial division?

 

[Makro]

f'(x)= (x²-4x)/(x-2)²
f'(1)= (1-4)/4
=-3/4
f'(-1)=(1+4)/4
=5/4
therefore maximum

Right I got that, but for the other root, I got:

f'(5) = 25-20/9
= 5/9

Leaving it positive on both sides for the root 4. Hi? and help? thanks

 

[tommykins]

thanks again.
i understand what to look for with algebraic manipulation
but what are general results for oblique assymptoes for polynomial division?

most of the time it's just a simply line, (ie. y = x+4) but SOMETIMES (rarely) they can give you a hyperbola or a parabolic assymptote.

 

[tacogym27101990]

Right I got that, but for the other root, I got:

f'(5) = 25-20/9
= 5/9

Leaving it positive on both sides for the root 4. Hi? and help? thanks

you need to remember that there is a discontinuity at x=2
so testing x=3
f'(3) = 9-12/q
=-3
hence minimum.