Question

Sketch the Curve: y=x^4-8x^3+18x^2-8x-3

Answer 1

do you know calculus..?

Answer 2

Refer to the attached plot.

Answer 3

oh thanks., are you using a software or some sites that graphs the curve?

Answer 4

Using Mathematica v9 Home Edition, a very large and powerful Mathematics software system.

Answer 5

do I need to install it? cause I need to graph two more

Answer 6

and I also need to prove it at the same time but the graph is a great help to check if my answer is right

Answer 7

I believe the installation media is 1.5Gb
I can plot the two expressions for you now.

Answer 8

wow thanks

Answer 9

[4y=20x ^{3}-3x ^{5}\]

Answer 10

y=x^5-5x^4+20x^2

Answer 11

4 y = 20 x^3 - 3 x^5

Answer 12

y=x^5-5x^4+20x^2

Answer 13

thank you very much!!!

Answer 14

God bless now I'll do the solving ahahah

Answer 15

Your welcome.

Answer 16

\[ \large y=x^4-8x^3+18x^2-8x-3\]
First, the ends. This is a fourth degree polynomial, which is to say an even power of x is the most significant term. When x gets very large in the positive direction \( x^4\) will get much larger than the other terms and will be positive. When x gets very large in the negative direction, \(x^4\) will again be more significant then the other terms, and will be positive since the power (4) is even. So we know the function swoops down from high in the second quadrant, does something, and then goes back up.

Now let's look for the zeros.

\[ \large y=x^4-8x^3+18x^2-8x-3\]

The coefficient on \(x^4\) is 1, and the constant term is -3. So we should suspect a factor is either \((x + 3))\) or \((x-3)\). If you are lucky, you try \((x-3)\) first.

Sure enough, \((x - 3)\) divides in easily.

\[ \large y=(x-3)(x^3-5x^2+3x+1)\]

Now looking at \((x^3-5x^2+3x+1)\) we see the coefficient on \(x^3\) is 1 and the constant term is 1. So we suspect another factor is \((x+1)\) or \((x-1)\). If you are lucky you try \((x-1)\) first.

\[ \large y=(x-3)(x-1)(x^2-4x-1)\]

At a glance you can see this won't factor, so you have to complete the square with \((x^2-4x-1)\)

\[(x^2-4x-1) -5 = (x^2-4x+4) -5= (x-2)^2 -5 \]

So the other 2 factors are

\[ x = 2\pm\sqrt 5\]

\[ \large y=(x-3)(x-1)(x-2+\sqrt 5)(x-2-\sqrt 5) \]

And the 4 zeros are \(3.\;1,\;2+\sqrt 5,\;2-\sqrt 5 \)

Now if you are doing calculus, you take the first derivative to find the critical points.

Otherwise logic tells you it swoops down, and crosses the x-axis at \(2-\sqrt 5\). It next crosses the x-axis at 1, so it has to go back up to do that. Then back down to cross the x-axis at 3, and back up to cross at \(2+\sqrt 5) and then takes off in the positive direction.

Answer 17

yes, we have the same answer., thanks :)

Answer 18

\[4y = 20x^3 - 3x^5={1 \over 4}(20x^3-3x^5)={1 \over 4}x^3(20-3x^2)=(-1){1 \over 4}x^3(x-\sqrt{20 \over 3})(x+\sqrt{20 \over 3})\]

The ends: \(x^5\) is the most significant term. Since three is an odd power \(x^5\) is a very large negative number when x is a large negative number, but the term has a negative coefficient, so on the left end, the graph gets to be a large positive number. But the same logic, as x gets to be a very large positive number, y become a large negative number.

There are three degenerate roots at \(x=0\). There is no constant term so we know this is an odd function ( \( f(-x) = -f(x)\) ). There are zeros at \(\sqrt{20 \over 3}\) and \( -\sqrt{20 \over 3}\).

So it swoops down, crosses the x-axis at the negative irrational root, comes back up to cross the x-axis at zero, comes down to cross the x-axis at the positive irrational root, and then continues to plunge.