Thread: Further Maths

1a) Find the set of values of x for which 0 < ( 2 - x - x^2 ) / x^2 < 2

1b) Find the cordinates of the turning point on the curve y = ( 2 - x - x^2 ) / x2

2) sketch y = root of (x-2)

3) Find x such that:
a) x^4 - 5x^2 6 > 0
b) | 3 - 2x < or equal to | x + 4 |

4a) show y = x - 6 is an asymptotoe of the curve y = (x-1)(x-3)/(x+2)

b) Sketch the curve showing cleanly the other asumptote, the points of intersection of the curve with the co-ordinate axes.

PLEASE PLEASE PLEASE HELP

Originally Posted by ziad123

1a) Find the set of values of x for which 0 < ( 2 - x - x^2 ) / x^2 < 2

1b) Find the cordinates of the turning point on the curve y = ( 2 - x - x^2 ) / x2

Mr F says: Solve . You can get dy/dx using either the quotient rule or seeing that .

2) sketch y = root of (x-2) Mr F says: Upper half of sideways parabola. Vertex at (2, 0).

3) Find x such that:
a) x^4 - 5x^2 6 > 0

Mr F says: Do you mean x^4 - 5x^2 + 6 > 0 or x^4 - 5x^2 - 6 > 0? Either way, I suggest drawing a graph and looking for the values of x for which the graph is above the x-axis.

b) | 3 - 2x < or equal to | x + 4 | Mr F says: You might start by drawing graphs of y = |3 - 2x| = |2x - 3| and y = |x + 4|. Then find the x-coordinates of the intersection points ....

4a) show y = x - 6 is an asymptotoe of the curve y = (x-1)(x-3)/(x+2)

Mr F says: using polynomial long division .....

b) Sketch the curve showing cleanly the other asumptote, the points of intersection of the curve with the co-ordinate axes. Mr F says: Vertical asymptote x = -2. You should know how to find x- and y-intercepts (especially y-intercept!!).

PLEASE PLEASE PLEASE HELP

1a) has been answered here: http://www.mathhelpforum.com/math-he...lp-urgent.html

It would help if you showed some of your work and said where you were stuck.