curve sketching/further graphs questions (1 Viewer)

[gc653]

Hey all

1. To sketch this graph

y= (x^2 -3x+2)/(x-3)

how do you know if your curve has to be drawn inside or outside the oblique/horizontal/vertical asymptotes??

2. Also say if denominator has a + sign instead of - (so that there is no vertical asymptote) how do you know where to draw weird curve that has a bump within the straight line??

E.g with this graph

y= (2x^3 +2x + 3)/ (x^2 + 1)


Thanx

 

[Crisium]

Hey all

1. To sketch this graph

y= (x^2 -3x+2)/(x-3)

how do you know if your curve has to be drawn inside or outside the oblique/horizontal/vertical asymptotes??

2. Also say if denominator was x + 3 instead of x-3 how do you know where to draw weird curve that has a bump within the straight line??

E.g with this graph

y= (2x^3 +2x + 3)/ (x^2 + 1)


Thanx

Vertical Asymptote: Equate the denominator to zero.

Horizontal Asymptote: Do the limit of x approaches infinity. When the highest power is on top it usually means it's an oblique asymptote.

"Bumps": Use the first derivative test to find stationary points and determine their nature using the neighbourhood check or the second derivative check. If you want to be even more detailed then find the points of inflection using the second derivative.

 

[Crisium]

https://www.desmos.com/calculator

^ Use this website if you want to see what it looks like

 

[braintic]

Hey all

1. how do you know if your curve has to be drawn inside or outside the oblique/horizontal/vertical asymptotes??

2. Also say if denominator was x + 3 instead of x-3 how do you know where to draw weird curve that has a bump within the straight line??

E.g with this graph

y= (2x^3 +2x + 3)/ (x^2 + 1)

1. Just substitute a large value of x to see which side it lies on.
BTW, horizontal/oblique asymptotes CAN be crossed (though not in this question).

2. Confused, since you talk about a denominator of x+3, then use x^2 + 1

 

[Carrotsticks]

Hey all

1. To sketch this graph

y= (x^2 -3x+2)/(x-3)

how do you know if your curve has to be drawn inside or outside the oblique/horizontal/vertical asymptotes??

2. Also say if denominator was x + 3 instead of x-3 how do you know where to draw weird curve that has a bump within the straight line??

E.g with this graph

y= (2x^3 +2x + 3)/ (x^2 + 1)


Thanx

For your first point, decompose the numerator to obtain x + 2/(x-3). When x is larger than 3, we've got y=x+(positive number) so it'll always be greater than y=x beyond x=3, hence it approaches y=x from above.

 

[gc653]

1. Just substitute a large value of x to see which side it lies on.
BTW, horizontal/oblique asymptotes CAN be crossed (though not in this question).

2. Confused, since you talk about a denominator of x+3, then use x^2 + 1

Ye sorry just ignore the x + 3 !!!!!!! I haven't been taught derivatives yet but is that the only way to find where the 'bump' starts and peaks on the line????

 

[Crisium]

Ye sorry just ignore the x + 3 !!!!!!! I haven't been taught derivatives yet but is that the only way to find where the 'bump' starts and peaks on the line????

Pretty much

Since you're in year 11 though, I doubt they will expect you know that in the exam so once you put in the asymptotes, substitute points into the equation to determine the nature of the graph on certain areas of the number plane; from this you will be given a rough idea of the shape of the graph