# Sketch a graph:

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It seems as if no-one wants to answer my questions :(

07 October 2013

Hi lala, Do you know how to differentiate a function? i.e. how to find y'

08 October 2013

2013-10-08_11-17-58_506.jpg

08 October 2013

sorry! upside down image. I'll try again..

08 October 2013

2013-10-08_11-17-58_506.jpg

08 October 2013

Thanks for replying back Ferdo! :))I don't understand how you got 2 and -2 for where the graph crosses the x-axis, isn't it just (2,0)Also, yeah I know how to find y, its 1 and not zero ( it was a typo)

08 October 2013

Also, can you help me with these questions... I'm struggling really badly with fm as...1) show that the curve y= (x^2+2x+2)/(x^2-x-1) exists only when y is less than or equal to -2 and for y is greater than or equal to 0.4. Hence find the coordinates of the stationary points of the curve.2)a) a curve has equation y=(x+1)/(x)(x-3)Show that there are no values of x for which -1<y<-1/9 B) hence determine the coordinates of any stationary points of the curve.C) state the equations of any asymptotes of the curve.3) a curve has equation y= (x^2)/(x^2+3x+3)A) prove that for all real values of x, y satisfies the inequality 0 is less than or equal to y and y is greater than or equal to y ( don't know how to write the symbols on keyboard)B) hence find the coordinates of the stationary points of the curveC) find the equations of any asymptotes of the curveCan you please help me ? And can you explain how to do the questions an I really need an answer by tomorrow morning....I really need to know how to do it!!!! Please please answer as soon as you are able toThanks! :)

08 October 2013

Lots of questions , i'll just do one :-Q2 and -2 comes from the values of x that make (x2-4) zero.Q1. stationary points are the points like "Max" and "min" in my drawing earlier. if you plot the function, you'll see that the curve is always above and below those values.Q2. You really need to know how to "differentiate". it's just a mathematical operation with functions. then it allows you to calculate Max and min in a curve, and answer questions like this.Other way to do it would be with "inequalities" , I mean ">" and "<" equations.For instance show that -1<(x+1)/(x)(x-3) and (x+1)/(x)(x-3)<-1/9 cannot be solved at the same time for any xQ3. Similar to the other questions, don't be so cheeky

08 October 2013

Okay, differentiation is taught a bit later. Hope a bit of help with Q2 still is useful.Q2. -1<(x+1)/(x)(x-3) ; -(x)(x-3)<(x+1) ;3-x2<x+1 ;3<x2+x+1 ;0<x2+x-2 : solving 0=, the two solutions for x are 1 and (-2)For the other values of x that are not 1 and (-2), let see quickly:x=-10 for example : x2+x-2 is very positive (88)x=0 : x2+x-2 is negative (-2)x=10 : x2+x-2 is positive (108) So, drawn in the LINE OF X, to see it clearly at a glance: see whitrboard next

10 October 2013

sketch-3.svg

10 October 2013

I really don't like this whiteboard :-(

10 October 2013

sketch-4.svg

10 October 2013

So, for all the values of x, 0<x2+x-2 is true when x is in the range of values less than (-2) and more than 1. Same is true for 0<(x+1)/(x)(x-3), because it's the same expression written different.That solves the first inequality -1<(x+1)/(x)(x-3) or -1<y (same thing)Still have to do (x+1)/(x)(x-3)<-1/9 or y<-1/9, put the solutions in a line of the x, together with the other line drawn, and see if there is any value of x that.... makes both expressions true.You find that there aren't any. Which means no y takes the value between-1 and -1/9 but can take other values outside of that range.it follows that y has Max and min in y=-1 and y+-1/9We did the first one, when y=-1, x is either -2 or 1, sooothe coordinates of those "stationary points" are (-2,-1) and (1,-1) Ta-daa . Well I'm going to draw it which is more fun than writing.

10 October 2013

graph_thursday.jpg

10 October 2013

irvek89mlk_3.png

13 February 2014

sketch-7.svg

13 February 2014

See graphical solution below. :D

13 February 2014

daum_equation_1392308939691_th_2.png

13 February 2014

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