# Math Help - graph

1. ## graph

hi, i've got another question on graphs, its part of a worked example, so it actually gives the answer but i don't understand:

sketch the graph of $y = (2x + 3)(x - 1)^2$

the coefficient of $x^3$ is +2 so the shape of the graph is the same $y = x^3$ when x is numerically very large (this i understand). When $x = 0, y = (3)(-1)^2 = 3$ so the graph crosses the y- axis at 3 (this i understand also). When $y = 0, (2x + 3)(x - 1)^2 = 0$ so $x = \frac {3}{2} -$ and $x = 1$ (repeated root) so the graph crosses the x-axis at $x = \frac {3}{2} -$ and touches the x-axis at $x = 1$. When $x = -1, y = (1)(-2)^2 = 4 > 3$ so the graph must turn in the second quadrant.

the last two points i don't understand, could someone tell me how you would tell if the graph crosses or only touches the x-axis and how you would tell by using x = -1 or x = + 1 or whatever that it turns in the second quadrant?

thanks for any help, mark

ps i tried putting the minus sign on the left for the fractions but it didn't work, i'm going to try to figure out how its done now

2. Originally Posted by mark
hi, i've got another question on graphs, its part of a worked example, so it actually gives the answer but i don't understand:

sketch the graph of $y = (2x + 3)(x - 1)^2$

the coefficient of $x^3$ is +2 so the shape of the graph is the same $y = x^3$ when x is numerically very large (this i understand). When $x = 0, y = (3)(-1)^2 = 3$ so the graph crosses the y- axis at 3 (this i understand also). When $y = 0, (2x + 3)(x - 1)^2 = 0$ so $x = \frac {3}{2} -$ and $x = 1$ (repeated root) so the graph crosses the x-axis at $x = \frac {3}{2} -$ and touches the x-axis at $x = 1$. When $x = 1, y = (1)(-2)^2 = 4 > 3$ so the graph must turn in the second quadrant.

the last two points i don't understand, could someone tell me how you would tell if the graph crosses or only touches the x-axis and how you would tell by using x = -1 or x = + 1 or whatever that it turns in the second quadrant?

thanks for any help, mark

ps i tried putting the minus sign on the left for the fractions but it didn't work, i'm going to try to figure out how its done now
Since the (x-1) term is squared, x= 1 is a double root of the equation. That's why the graph touches rather than crossing at x= 1. Another way to see that is note that near x= 1, 2x+ 3 is always positive so the sign of $(2x+3)(x-1)^2$ depends only on $(x-1)^2$. But that is squared so is always postive. It doesn't matter whether x< 1 or x> 2, the function is still positive. Around x=-3/2, $(x-1)^2$ is always positive so the sign of the function depends on (2x-3) which is negative for x< -3/2 and positive for x> -3/2. So the sign of the function changes from negative to positive: the graph crosses the axis at x= -3/2.

"When $x = 1, y = (1)(-2)^2 = 4 > 3$ so the graph must turn in the second quadrant. " is wrong. when x= 1, y= 0. You mean "when x= -1". The value of the function when x= -3/2 is 0, when x= -1 is 4, and when x= 0 is 3. That's what they mean by "turn". The function has gone up from 0 to 4 and then back down to 3 between -3/2 and 0.

3. when you say x = 1 is a double root of the equation do you mean the same as "it has two distinct real roots"? because i've learnt that it crosses the x-axis twice when its bigger than 0. i've probably got it mixed up though. i think i understand the stuff about the graph turning now. i'm just confused about this touching the axis crossing the axis business

4. Have you started calculus yet.. like differentiation?

If the discriminant b2 - 4ac equals zero, the radical in the quadratic formula becomes zero.

In this case the roots are equal; such roots are sometimes called double roots.

This applies to your case too.

5. no i haven't. one thing i've noticed about this book is that as well as not explaining things well, some things that are really useful to know for certain subjects are just left out till later. why? do you think calculus/differentiation would help?

6. Calculus helps enormously when trying to identify turning points on cubic graphs. I don't want to sidetrack the topic too much, but:

$y=2x^3-x^2-4x+3$

This represents the gradient at any point on the graph:
$\frac {dy}{dx}=6x^2-2x-4$

Obviously, the gradient at a turning point is 0:
$0=6x^2-2x-4$
$0=2(x-1)(3x+2)$

So clearly, a turning point exists at x=1 and x=-(2/3)

But nevermind, this is completely meaningless and irrelevant to you! Is there anything else you don't understand from the worked example?

7. ha thanks, yeah i don't understand what that means entirely, i noticed that you were factorising though. another thing in this book i don't understand is that one of the questions is sketch the graph of $y = x^3 - 1$ and another is $y = (x - 1)^3$. aren't they the same? the graph for the first crosses straight over the x axis whereas the second it lays flat on it then crosses properly, why is this?

8. Nooo.. This is a common mistake in mathematics.

is not the same as

actually equals:
$y=(x-1)(x-1)(x-1)$
$y=x^3-3x^2+3x-1$

That's an entirely different graph to

9. ah yeah, that was a stupid mistake. i still wouldn't know how to draw the graph though.

10. Actually, I remember a cute little mnemonic to assist in sketching graphs when calculus hasn't been taught.

Cubic polynomials, that aren't in the standard form, always tend to have 2 turning points... a hump (maximum point) and a U (minimum point).

Mneomonic:
When the x^3 term is positive, the camel's hump is on the left.

When the x^3 term is negative (eg. -x^3) the camel's hump is on the right.

Lol, in retrospect it's actually rather pathetic.

11. lol sounds pretty good, i understand the turning point stuff now. i just don't get how you can tell whether it crosses or touches the x-axis. no doubt i will in time though because i've managed to understand everything else so far. cheers for the help

12. Just keep it simple

If there's a squared it touches, if there's not it crosses.

EG.

(2x+3).. is it squared? No, so therefore it crosses this root (aka x-intercept).

(x-1)^2... is it squared? Yes, so therefore it touches the root (x-intercept).

13. that actually seems to work with some of these questions, i'll let you know how i get on