# Thread: Really Stuck, graphs of functions???

1. ## Really Stuck, graphs of functions???

The Question Reads:

draw the graphs of the functions listed over the given range's of x in steps of one unit. from these, estimate the nature of the roots.

(i) y = x(2) + 2x - 6 (Range of x = -5 to +3)

I understand pretty much none of this, i dont know how to draw the graph because i dont know what to plot, or how to calculate the data, in order to plot. I've got no literature, had no worked examples and a useless lecturer who likes to show how clever he is by means of baffling myself and the other 70 students with utter gobledegook.

anyway, enough about the poor quality of my lecturers. can anyone please help, i know its a very basic question but i really just dont know where to start. many thanks

2. Originally Posted by Parton-Bill
The Question Reads:

draw the graphs of the functions listed over the given range's of x in steps of one unit. from these, estimate the nature of the roots.

(i) y = x(2) + 2x - 6 (Range of x = -5 to +3)

I understand pretty much none of this, i dont know how to draw the graph because i dont know what to plot, or how to calculate the data, in order to plot. I've got no literature, had no worked examples and a useless lecturer who likes to show how clever he is by means of baffling myself and the other 70 students with utter gobledegook.

anyway, enough about the poor quality of my lecturers. can anyone please help, i know its a very basic question but i really just dont know where to start. many thanks

You just plug in a value of x from a range of -5 to +3. ie: -5, -4, -3, -2, -1, 0, 1, 2, 3 thts all you do then you will get the value of y, you should do at least 3 to 5 points

3. Originally Posted by Brent
You just plug in a value of x from a range of -5 to +3. ie: -5, -4, -3, -2, -1, 0, 1, 2, 3 thts all you do then you will get the value of y, you should do at least 3 to 5 points
I've plotted those as i understand from the above:

so i should have calc's such as -5(2) + (2 x -5) - 6 = -41
-3(2) + (2 x -3) - 6 = -21
3(2) + (2 x 3) - 6 = 9

Etc, this would give me what looks to be a straight line, or possibly curved graph, where as i had expected a negative curved graph such as:

Can i also ask, would i begin plotting at -5 on the negative X axis?

Many thanks

4. Originally Posted by Parton-Bill
I've plotted those as i understand from the above:

so i should have calc's such as -5(2) + (2 x -5) - 6 = -41
-3(2) + (2 x -3) - 6 = -21
3(2) + (2 x 3) - 6 = 9

Etc, this would give me what looks to be a straight line, or possibly curved graph, where as i had expected a negative curved graph such as:

Can i also ask, would i begin plotting at -5 on the negative X axis?

Many thanks

No u cant just plug -5 on the axis. u have to plug -5 in as the value of x then solve for y ie:

y = x(2) + 2x - 6
y=-5(2) + 2(-5) -6
y = -10 + -10 -6
y= -26 so first point could be (-5,-26)

Usually i just use the smallest values of x -2,-1,0,1,2 just so i can plot it on a piece of paper and look to see how the graph is gonna go.

5. Yep, with you on that, dont think i worded it very well.
What i meant was when x = -5, y = -41 so the co-ordinates for the first plot would be X = -5, and Y = -41, and so on???

One more question :0)

with an equation such as y = -6 + 4x - x^2, is it better to tranpose this, in order to bring x^2 to the front, eg: y = x^2 - 4x + (-6) ???

I'm not sure why but i feel with quadratic functions they should start x^2???

Thanks again Brent, really appreciate your time buddy