# line

• Jan 14th 2008, 04:58 PM
fabi
line
hi, im having trouble understandind what i have to do. ive tried reading over the text book even though ive only started chap1. i have no clue how to tackle this.the q is:

sketch the graph and state the range:

y=x+1; xE]2,infinty]

*note 1st bracket must be other way*
• Jan 14th 2008, 05:06 PM
mr fantastic
Quote:

Originally Posted by fabi
hi, im having trouble understandind what i have to do. ive tried reading over the text book even though ive only started chap1. i have no clue how to tackle this.the q is:

sketch the graph and state the range:

y=x+1; xE]2,infinty]

*note 1st bracket must be other way*

Do you know how to draw the line y = x + 1. Draw it in pencil.

Note that when x = 2, y = 3 (why?). Mark the point (2, 3) with a small coloured-in circle (called a closed circle) (do you see where this point comes from?). Do this in pen. This circle means that the point is included.

From the point (2, 3) onwards, ink the line with a pen.

Rub out the pencil bit that's left.

The range is all the y values. Look at the graph. Do you see that $3 \leq y < \infty$? That's the range.
• Jan 14th 2008, 06:02 PM
fabi
another q
thanks that made it so easy.
another q.

find for f(x)=3x^2+x-2

a. x:f(x)=0
• Jan 14th 2008, 06:19 PM
fabi
thanks dw i worked it out myself
• Jan 14th 2008, 10:55 PM
fabi
another
the question states sketch the graph with the following rule

g(x)= (x^2+5 x>0
5-x -3<x<0
8 x<-3

*note the bracket is for all 3 on the left side, one big bracket*
5-x is the eq and 8 is the next
• Jan 15th 2008, 12:21 AM
earboth
Quote:

Originally Posted by fabi
the question states sketch the graph with the following rule

g(x)= (x^2+5 x>0
5-x -3<x<0
8 x<-3

*note the bracket is for all 3 on the left side, one big bracket*
5-x is the eq and 8 is the next

Hello,

if I understand your question correctly you want to draw the graph of

$g(x)=\left \{ \begin{array}{ccc}8&\text{ if }&x<-3\\5-x&\text{ if }&-30 \end{array}\right.$

According to the notation of the function there are gaps in the graph at x = -3, x = 0 because g(-3) and g(0) don't exist.

PS: If you have a new question start a new thread. Otherwise you risk that nobody will notice that you want some help.