A few "tricky" (probably easy) questions... (1 Viewer)

[davidbarnes]

Can anyone tell me how i woudl solve the following, which have got me stumped. Any help is needed as the exam is soon. Thankyou.

 

[watatank]

first one. I *think* you gotta factorise (difference of two cubes) and then cancel out the denominator. then replace the variable with the 5's and youre set. that might be wrong though.

second one you just factorise the numerator, gets rid of the h in denominator. then you just replace the 'h's with 0's

third one: its a hyperbolic function, because its a fraction. find vertical assymptotes (defines where x doesnt exist, hint- look at denominator) then find horizontal assymptotes (limit as x approaches infinity). test a few points and you should be able to draw your curve

fourth one: distance formula.

D^2 = (x₂-x₁)² + (y₂-y₁)²

you know the distance. you've got all points bar one so solve for the final one.

fifth: put in the form y = mx + b and m's your gradient.

 

[Evergreen]

we havent got into much detail with limits yet

so ill just answer last ones

4. d=sqrt [(a-3)^2 +(-1-4)^2]=5

.: (a-3)^2 +25 =25

a=3

5. just multiply everything by the denominators you'll get

3x+8y+24=0

y=-3/8x -3

.: dy/dx= -3/8

1. i got this question lol

(y-5)(y^2 +5y +25)/y-5

sub in 5 into (y^2 +5y +25) and the expression=75.

 

[watatank]

dy/dx is the gradient function of x, not the gradient itself.

although if you differentiate y with respect to x it is true that the gradient will be -3/8 for all x (assuming you did the calculation right)

 

[jb_nc]

1) 125 is 5³. watatank was right. factorise then cancel out the denominator.

for 5) I would just set the equation equal to y then find dy/dx whatever is left is the gradient.

 

[darkliight]

1. lim_{y->5} (y^3 - 125)/(y-5) = lim_{y->5} (y-5)(y^2 + 5y + 25)/(y-5) = lim_{y->5} (y^2 + 5y + 25) = ...

2. You have a h in every term in the numerator, so the h in the denominator cancels with one h in every term in your numerator. This leaves you with ... sub in h = 0 ...

3. y = (x^2 + 3x)/(x + 3) = x(x+3)/(x+3) = x ... graph me.

4. Using the distance formula, 5^2 = (3 - a)^2 + (4 - -1)^2, so 0 = (3-a)^2, that is, a = 3.

5. This is the equation of a line (ax + by + c = 0). Write it in the form y = mx + b, then the gradient is just m. You can use differentiation, but, it's a bit overkill I think.

 

[davidbarnes]

Thanks for all the excellent help guys. A couple of those seems really easy now (distance formula, how the heck did I not think of that?, lol).

I'm still not really sure on No.3 though.

 

[watatank]

3. y = (x^2 + 3x)/(x + 3) = x(x+3)/(x+3) = x ... graph me.

is this valid? if it is then you just gotta remember x =/= -3 (can't divide by zero) so its a discontinuous line with no solution where x = -3.

 

[jb_nc]

The lim_x->+inf = inf/inf

Use L'Hopitals or cancel out like terms.