# Graphs

• Dec 10th 2007, 09:30 AM
Chez_
Graphs
Hey, i was asked to find the co-ordinates of the points were the graph, y=x^2-7x+6 crosses the axis, to find this i did x =1,2,3-1,-2 etc and found the y value. i got that (0,6)(1,0)(6,0) crossed, but this looks a little strnage to draw a curve on the graph.... have i done it right?

Also what would be the best way of working out y=x(x-3)^2, so that i could sketch it?

Thank you!
• Dec 10th 2007, 09:35 AM
topsquark
Quote:

Originally Posted by Chez_
Hey, i was asked to find the co-ordinates of the points were the graph, y=x^2-7x+6 crosses the axis, to find this i did x =1,2,3-1,-2 etc and found the y value. i got that (0,6)(1,0)(6,0) crossed, but this looks a little strnage to draw a curve on the graph.... have i done it right?

"...crosses the axis..." Not very specific. Given the rest of the post I imagine you are looking for both the x and y intercepts?

-Dan
• Dec 10th 2007, 09:36 AM
janvdl
Quote:

Originally Posted by Chez_
Hey, i was asked to find the co-ordinates of the points were the graph, y=x^2-7x+6 crosses the axis, to find this i did x =1,2,3-1,-2 etc and found the y value. i got that (0,6)(1,0)(6,0) crossed, but this looks a little strnage to draw a curve on the graph.... have i done it right?

Also what would be the best way of working out y=x(x-3)^2, so that i could sketch it?

Thank you!

That graph will only have two x-intercepts...

Find the values of x where y = 0

\$\displaystyle 0 = x^2 - 7x + 6\$

\$\displaystyle 0 = (x - 6)(x - 1)\$

\$\displaystyle So \ x = 6 \ or \ x = 1\$
• Dec 10th 2007, 09:37 AM
janvdl
Quote:

Originally Posted by topsquark
...other x intercept is (-6, 0).

-Dan

(+6;0) Topsquark?
• Dec 10th 2007, 09:40 AM
topsquark
Quote:

Originally Posted by Chez_
Also what would be the best way of working out y=x(x-3)^2, so that i could sketch it?

The best way is to do it point by point, just like you did the other one. You know that the x intercepts are (0, 0) and (3, 0), and you know the y intercept is (0, 0). Now plot the points for, say x = -1, 1, 2, 4, and 5. Now connect the dots with a smooth curve.

-Dan
• Dec 10th 2007, 09:42 AM
topsquark
Quote:

Originally Posted by janvdl
(+6;0) Topsquark?

I think I'm having one of those days where I'm having one of those days. (Headbang) (Thank you.)

-Dan
• Dec 10th 2007, 09:44 AM
janvdl
Quote:

Originally Posted by topsquark
I think I'm having one of those days where I'm having one of those days. (Headbang) (Thank you.)

-Dan

Hehe, no problem. It happens to me a lot too. And it's usually the reason why i lose marks in exams or tests.
• Dec 10th 2007, 10:04 AM
Chez_
eq of lines, graphs
I have a picture of a line AB tht eq is y=4x-5 and passes through the point 9(2,3). The line BC is perp to AB and cuts the x axis at C. I need tofind the equation of line BC, which i have done, but i also need to find the x coordinate of C, but im not sure how do it? thnx

Also how could i express 3(in a square root thing)
--------
6-3(3 again in sqr root thing)
in the form of p+q 3(in sqroot)

Iv now with the help of you great people being able to complete my practise paper and practised and found some new tips lol thnx
• Dec 10th 2007, 10:13 AM
topsquark
Quote:

Originally Posted by Chez_
I have a picture of a line AB tht eq is y=4x-5 and passes through the point 9(2,3). The line BC is perp to AB and cuts the x axis at C. I need tofind the equation of line BC, which i have done, but i also need to find the x coordinate of C, but im not sure how do it? thnx

Also how could i express 3(in a square root thing)
--------
6-3(3 again in sqr root thing)
in the form of p+q 3(in sqroot)

Iv now with the help of you great people being able to complete my practise paper and practised and found some new tips lol thnx