Graph question

B

Bobby Jones

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I have a graph in front of me with no scale or numbers. It crosses the X,Y axis at 0,0,(O) oes up to a curve (Q), down to touch the X axis (P) then up into the air.

The equation is y = x^3 - 6x^2 +9x

The curve touches at the x -xis at the point P and has a maxima at point Q

a) Show the the equation of the curve may be written in the form y= x(x-a)^2 where a is a constant. Hence state the co-ordinates of the point P.

My answer: I got a as -3, but I cant find the co-ordinates for P, it touches the x axis so x = 0, what is Y (0,?)

b)Find the gradient of y and hence determine the point Q

My answer: I differentiated the equation to make dy/dx = 3x^2 - 12x + 9

c) Calculate the area under the graph for the region bounded by the origin O and point P.

My answer: I intergrated the equation to make 1/4 (x^4) - 2x^3 + 9/2 (x^2) + C


Am I right or wrong?
 
y = x^3 -6x^2 + 9x = x(x^2 - 6x + 9) = x(x-3)^2

a) How did you get a = -3, which I'm guessing you means, rather than a = 3?

(3,0) -- What else would it be? If it is on the x-axis, then y = 0. You seem a bit twisted about.

b) x = 3 is out of the question. You have the first derivative, now what good does it do you?

c) You'll have to know Q, won't you? You did not integrate any equation. That makes no sense at all. You appear to have found an anti-derivative. I don't see an integral in there, anywhere.

Correced Sign in a) and exponent at the beginning. Sleep typing again.
 
tkhunny,

I did get -3 for a, if you look at my original post.

How did you get

y = x^3 -6x^2 + 9x = x(x^2 - 6x + 9) = x(x-3)^3

Surely ,it should be ^2

How do i find Q, do i differentiate the equation?
 


Bobby Jones said:
I did get -3 for a, if you look at my original post.

a = 3, not -3

Otherwise, (x - a)^2 would be (x - (-3))^2 = (x + 3)^2


How do i find Q, do i differentiate the equation?

Yes. Maybe you missed an important point in your lessons.

You found dy/dx. That quadratic polynomial represents the slope of the tangent line at any point on your graph, when you evaluate it using the x-coordinate at that point.

Do you know what the slope value is for any line that is tangent to a graph at a local maximum or local minimum point?
Click to expand...

And, yes, it's clear that tkhunny meant to type ^2

 
Do you know what the slope value is for any line that is tangent to a graph at a local maximum or local minimum point?
Click to expand...

Is it y= mx+c ?

So I differeniated the original expression y = x^3 -6x^2 + 9x

to make dy/dx = 3x^2 - 12x + 9 (which i believe is the gradient)

You found dy/dx. That quadratic polynomial represents the slope of the tangent line at any point on your graph, when you evaluate it using the x-coordinate at that point.
Click to expand...

So would I take 3x^2 - 12x + 9 and factorise it to (3x -9)(x-1) to then give me x= 3 and x= 1 ? I'm getting stuck.

If I put put either value it comes to zero. So how can I work out the coordinates of the maxima Q from differentiating the original equation?
 
Bobby Jones said:
I have worked out Q's co-ordinates to be (1,4)

Is this correct?
Click to expand...

Plot the function within the domain[-1,4].

That will answer your questions.
 
Subhotosh Khan

I think your wrong, putting x = -1 when y = 4 into the function dosent work.

Q's co-ordinates are (1,4)

I differenciated the original function. Then took the factors of that; (3x-9)(x-1), therefore x is 3, or 1. And P is already at point 3 on the x axis therefore Q must be at 1.
 
You need to review pre-calc concepts - before attempting these advanced problems.

I said choose DOMAIN to be [-1,4] - that means for the graph, use - 1 as MINIMUM 'x' and '4' as MAXIMUM 'x'.
 
Cheers guys, sorry for the confusion along the way. I plugged used the domain [-1,4] and it came out correct.

Now how do I find the "trapped" area between the hump and the x axis, Do i intergrate it once? Then thats my answer.


Therefore

y= x^3 - 6x^2 + 9x

= (1/4)x^4 - 2x^3 + (9/2)x^2 + C units/squared
 
Bobby Jones said:
Cheers guys, sorry for the confusion along the way. I plugged used the domain [-1,4] and it came out correct.

Now how do I find the "trapped" between the hump and the x axis, Do i intergrate it once? Then thats my answer.


Therefore

y= x^3 - 6x^2 + 9x

= (1/4)x^4 - 2x^3 + (9/2)x^2 + C units/squared
Click to expand...

Have you learned "definite integrals"?
 
No I havent leant them yet. I take it I need them to work out the area under the curve.

Are they the curly looking d. Give me a quick introduction/lesson on how they are applied please subhotosh khan
 
Bobby Jones said:
No I havent leant them yet. I take it I need them to work out the area under the curve.

Are they the curly looking d. Give me a quick introduction/lesson on how they are applied please subhotosh khan
Click to expand...
It's an old-fashioned S standing for infinite "sum," not a d.

First, you need to understand the anti-derivative (also frequently called the indefinite integral).
Definition: If and only if G(x) = F'(x), then \(\displaystyle \int\)G(x)dx = F(x) + k, where k is any real number.
A lot of what you learn in integral calculus is how to find the anti-derivative of various functions.
Sums are easy. If G(x) = H'(x) + J'(x), then \(\displaystyle {\int}\)G(x)dx = H(x) + J(x) + k.

Powers are easy too. If G(x) = ax[sup:1svhrsng]n[/sup:1svhrsng], then \(\displaystyle \int\)G(x) = [ax[sup:1svhrsng]n+1[/sup:1svhrsng]/(n + 1)] + k.

So polynomials, which are just sums of powers, are easy too.

A definite integral of F(x) gives the area between two vertical lines, the graph describing F(x), and the x-axis.
If the vertical lines are at p and q, with p < q, then
the area = \(\displaystyle \int\)G(x = q)dx - \(\displaystyle \int\)G(x = p)dx = \(\displaystyle \int_p_q\)G(x)dx = the definite integral. Notice that k disappears.
 
I get it, I take the intregral of the equation then apply the definite limits of where the curve intersects the x axis, which in this case will be 0 and 3.

Therefore

y = x^3 - 6x^2 +9x

intergrates to:

1/4 (x^4) - 2x^3 + 9/2 (x^2) between 0 and 3

1/4 (3^4) - 2x3^3 + 9/2 (3^2) = 20.25 - 54 + 40.5 = 6.75

1/4 (0^4) - 2x0^3 + 9/2 (0^2) = 0

6.75 - 0 = 6.75

Therefore the answer to the area under the curve is 6.75 units squared.

Have I got it right guys!
 
Bobby Jones said:
I get it, I take the indefinite intregral of the equation then evaluate it (or calculate the definite integral) where the curve intersects the x axis, which in this case will be 0 and 3. Yes that is what you do in this case. But the evaluation need not occur at points where the curve intersects the x-axis.

Therefore

y = x^3 - 6x^2 +9x

intergrates to:

1/4 (x^4) - 2x^3 + 9/2 (x^2) Actually the indefinite integral is (x[sup:25zj85lq]4[/sup:25zj85lq]/4) - 2x[sup:25zj85lq]3[/sup:25zj85lq] + (9x[sup:25zj85lq]2[/sup:25zj85lq]/2) + k everywhere

1/4 (3^4) - 2x3^3 + 9/2 (3^2) + k = 20.25 - 54 + 40.5 = 6.75 + k

1/4 (0^4) - 2x0^3 + 9/2 (0^2) + k = k

(6.75 + k) - k = 6.75

Therefore the answer to the area under the curve is 6.75 units squared. Looks good to me

Have I got it right guys!
Click to expand...
 
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