C1 Sketching Graphs

I think I know how to do it but I need to work the two x co-ordinates when they cut the x-axis and I just can't do it (I don't even know if it's possible).

The equation is y = x^2 -3x + 3.

I used quadratic equation but it comes out as a negative and you can't root a negative so what do I do?

Reply 1

Small123

15

kai4321

I think I know how to do it but I need to work the two x co-ordinates when they cut the x-axis and I just can't do it (I don't even know if it's possible).

The equation is y = x^2 -3x + 3.

I used quadratic equation but it comes out as a negative and you can't root a negative so what do I do?

Alright you got a negative root which is indeed correct. But what does this imply? :smile:

Reply 2

kai4321

OP

Small123

Alright you got a negative root which is indeed correct. But what does this imply? :smile:

I have no idea to be honest. I mean, it HAS to cut the x-axis because the other points make sense yet you can't work out x. My teacher made the equation up on the spot - basically I can't do it or I can?

Reply 3

veronica111

9

well if it doesnt have any roots, then it doesnt cross the x-axis, so put in, when x=0 to find out where it crosses the y-axis. So then you find out that it crosses the y axis at (0,3)

hope that helps :smile:

Reply 4

kai4321

OP

veronica111

well if it doesnt have any roots, then it doesnt cross the x-axis, so put in, when x=0 to find out where it crosses the y-axis. So then you find out that it crosses the y axis at (0,3)

hope that helps :smile:

Right but surely it can't get to (0,3) unless it does cut the x-axis?

Reply 5

Small123

15

kai4321

I have no idea to be honest. I mean, it HAS to cut the x-axis because the other points make sense yet you can't work out x. My teacher made the equation up on the spot - basically I can't do it or I can?

Complete the square of the equation:
y=x^2-3x+3
=(x-3/2)^2+3/4
gives positive parabola with minimum point (?,?) which is always positive hence it cannot =0.

Reply 6

veronica111

9

kai4321

Right but surely it can't get to (0,3) unless it does cut the x-axis?

yeah you can, look at this guys graph

<quote=zdo0o>(X,Y)
(0,3)

its like

....\...|.../
.....\_|_/
........|
_____|______
........|
........|
........|</quote>

btw I am VERY impressed with your graph drawing skills via keyboard aha :smile:

Reply 7

kai4321

OP

Oh sorry I made silly mistake of getting my axes mixed up. :frown:

Ok, right, well then you can do it but it doesn't cross - I think I've got it - thanks guys!!

Reply 8

KrazyKro

9

in future if u wanna know whether a graph crosses the x axis or not (number of solutions) than calculate the discriminant of the function by using b^2 - 4ac....

in this case...it will be (-3)^2 - 4 * 1 * 3 = -3....negative answer means no solutions....0 means 1 solution and positive answer means 2 solutions....

Reply 9

Leighthesim

2

an easy way to check whether it will cross the x-axis is to use the discriminant (the b^2-4ac part of the quadratic formula) if you have the cambridge Core 1 for edexcel maths text book (is green with orange patches on it) then it is on page 50.

Reply 10

JAKstriked

11

b^2-4ac

learn it. as has been said, greater than 0 implies 2 real roots, equal to 0 is 1 repeated root and less than zero (in this case) means no real roots (ie no intercept with the x axis)

C1 Sketching Graphs

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