Math Is Fun Forum

jigbobber

Member

Registered: 2008-08-13

Posts: 4

need more help =(

having more trouble with these..

1.Find the intercepts, asymptotes, stationary points and sketch the graph

2.show algebraically that

is a tangent to the curve

3. Find the value of a if the tangent to

at

passes through the origin

thanks if you can help again.

use to be able to do these but they're revision from the start of the year and im having trouble remembering

Last edited by jigbobber (2008-08-13 02:56:17)

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: need more help =(

1. What part are you having trouble with? For intercepts, set y = 0 and solve for x to find the x-intercepts, and set x = 0 to give the y intercept. For the asymptotes, factorise the bottom (difference of two squares). What values can x not take? For stationary points, differentiate using the quotient rule, set dy/dx = 0 and solve for x.

2. Set 4x - 2 = x^3 + x and bring everything onto one side: x^3 - 3x + 2 = 0. From here you should spot x=1 is a solution, so you know (x-1) is a root. You can use this to factorise the expression, giving (x-1)(x^2+x-2) = 0, which can then be factorised further to give (x-1)(x-1)(x+2). The repeated (x-1) shows that the two curves only touch at the point x = 1 but do not cross (therefore they are tangential to each other).

Last edited by Daniel123 (2008-08-13 03:07:51)

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: need more help =(

you missed a bit out there daniel, finding the values that x cannot take only gives you the asymptotes parralel to y-axis, you may also get an asymptote by taking the limit as x tends to positive/negative infinity, which may or may not be an asymptote parallel to x axis.

although from the orders of the polynomials you can see the asymptote in doing so will be parallel to x axis

asymptotes x = 1, x = -1

  asymptote y = 1

also for plotting, the best way is that once you have asymptotes, stationary points and intercepts is to get your graph paper, draw in the asymptotes with dotted lines, mark the stationary points and the nature of the stationary point (not always necessary, here you don't need to know it) and the intercepts, then, for asymptotes; evaluate the function from each side of asymptote to see in which direction the graph will go on that side of the asymptote.

Last edited by luca-deltodesco (2008-08-13 03:30:21)

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Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: need more help =(

Of course. That'll teach me to do maths when I've just woken up Thanks Luca.

jigbobber

Member

Registered: 2008-08-13

Posts: 4

Re: need more help =(

ah ok cheers for the help. any idea what to do with 3?

jigbobber

Member

Registered: 2008-08-13

Posts: 4

Re: need more help =(

anyone?

gnitsuk

Member

Registered: 2006-02-09

Posts: 121

Re: need more help =(

Hi,

Ok, so we have the curve:

So we differentiate this function to get the gradient function:

So when x = 2 the slope of the tangent line will be:

Also, we are told that when x = 2 this tangent line passes through the origin, therefore this line's equation is:

But this tangent line and the original curve are coincident at x = 2 therefore we have:

Solving this we arrive at:

See Result Graphically

Last edited by gnitsuk (2008-08-14 21:29:10)