finding X intercepts (1 Viewer)

[Zak Ambrose]

in regards to geometric application of the derivative.
i have no problem finding the x intercepts of a quadratic like
x^2 - 4x + 3 = y
i just factorise or use the general quadratic solution.
but how do i find the x intercepts or a cubic or even to the power of 4?

 

[vds700]

in regards to geometric application of the derivative.
i have no problem finding the x intercepts of a quadratic like
x^2 - 4x + 3 = y
i just factorise or use the general quadratic solution.
but how do i find the x intercepts or a cubic or even to the power of 4?

Just factorise it i guess.

e.g1. find the x-intercepts of the equation y = x³ - 27

make y = 0
x³ - 27 = 0
(x - 3)(x² + 3x + 9) = 0
therefore x = 3

e.g 2. Find the x-intercepts of the equation y = x⁴ - 49
again make y = 0
x⁴ - 49 =0
(x² + 7)(x² -7) = 0
x = +/- sqrt7

hope this helps

 

[Zak Ambrose]

it goes..
quadratic then cubic then quartic right?

 

[vds700]

it goes..
quadratic then cubic then quartic right?

yep

 

[Zak Ambrose]

how would i find the x intercepts on something like this..
f(x) = x^4 - 29x^2 + 100
let f(x) = 0
x^4 - 29x^2 + 100 = 0
x^4 - 29x^2 = -100
x^2(x^2 - 29) = -100
?

 

[Trebla]

Let u = x² and solve it like a quadratic. Don't forget to resubstitute u back into x to solve for the two solutions of x.

 

[Zak Ambrose]

and im left with the answers +- 2, +- 5.
thank you very much.

what about x^3 + x^2 - 5x - 6 = 0 ?
id assume id have to factorize.

wait dont worry, end up with complex roots or sumin.

what about x^4 - 2x^3 = 0
i would factorise and get x^3(x-2) = 0
x = 0 or 2, i know the answers right, just checking if my working is ok

 

[vds700]

and im left with the answers +- 2, +- 5.
thank you very much.

what about x^3 + x^2 - 5x - 6 = 0 ?
id assume id have to factorize.

wait dont worry, end up with complex roots or sumin.

what about x^4 - 2x^3 = 0
i would factorise and get x^3(x-2) = 0
x = 0 or 2, i know the answers right, just checking if my working is ok

yeah that is right

 

[addikaye03]

You will never be asked to fastorise an expression greater than degree 2 ( e.g x^2) unless theres a common term or something.. otherwise u will need to use a 3U result ( factor theorem).

 

[bored of sc]

how would i find the x intercepts on something like this..
f(x) = x^4 - 29x^2 + 100
let f(x) = 0
x^4 - 29x^2 + 100 = 0

You can do this:

Pretend it's a quadratic but just replace x² with x⁴ and x with x²

i.e. x^4 - 29x^2 + 100 = 0
What two numbers multiply to give 100 and add to give -29? -25 and -4
(x²-25)(x²-4) = 0

Now let each factor be equal to zero for values of x.

x²-25 = 0
x² = 25
x = +5

x²-4 = 0
x² = 4
x = +2

P.S Doesn't always work since it won't always be factorisable.

 

[Zak Ambrose]

cool.
last question.
y = (x-1)(x-3)^3
sketch after finding dy/dx, x intercepts, y intercept, stationary points.
i found the x intercepts, 1 and 3 obviously.
y intercept is 27.

im just having trouble find the stationary points, another cubic question.

stationary points occur for
f'(x) = 0
using product rule
3(x-1)(x-3) + (x-3)^3 = 0
expanding i get
4x^3 - 30x^2 + 72x - 54 = 0
where do i go from here?

 

[Chinmoku03]

I'm assuming you just forgot to add the square after the first (x - 3) after differentiating the whole thing, cos your resulting equation is correct.

You can try guessing values for x. Roots of the equation come in the form (Factor of constant) / (Factor of coefficient of leading term). In your case, it would be (Factor of 54) / (Factor of 4). Once you've found an x value that satifies the equation, you can then solve the remaining quadratic and hence find the stationary points.

 

[Zak Ambrose]

I'm assuming you just forgot to add the square after the first (x - 3) after differentiating the whole thing, cos your resulting equation is correct.

You can try guessing values for x. Roots of the equation come in the form (Factor of constant) / (Factor of coefficient of leading term). In your case, it would be (Factor of 54) / (Factor of 4). Once you've found an x value that satifies the equation, you can then solve the remaining quadratic and hence find the stationary points.

im not following sorry.

 

[Chinmoku03]

Ok, factors of 54 are: [1,2,3,6,9,18,27,54]. Factirs of 4 are: [1,2,4].

Let x be a fraction with a factor of 54 as its numerator and a factor of 4 as its denominator.

So I'll guess 1 for the numerator and 1 for the denominator, making x = 1. Sub x = 1 into the equation. You will see that x = 1 is not a solution to the equation. So now I'll try 2 as the numerator and 1 as the denominator, making x = 2. Rinse and repeat until you find an x that satisfies the equation. Once you've found a value for x, all that's left is a quadratic, which you know how to solve.

 

[Zak Ambrose]

oh ok.
so ill just sub in
4[1^3] - 30[1^2] + 72[1] - 54 = -8 so 1 isn't right
then ill sub in 2
4[2^3] - 30[2^2] + 72[2] - 54 = 2
is this method right? and if so? where's the quadratic left?

 

[Chinmoku03]

Yeah, and you keep going until you find a value for x that works. Don't forget that the x value could be a fraction. (ie 1/2.) And I forgot to mention this before, but it's possible for the numerator to be the negative value of a factor. So because 54 has [1,2,3,6,9,18,27,54] as factors, [-1,-2,-3,-6,-9,-18,-27,-54] are possible values for the numerator as well.

The quadratic that's left depends on the x value you find that works. I'll edit this post once I find an x value that works.

EDIT: Ok, the x value that works is x = 3. So:
4x³ - 30x² + 72x - 54 = (x - 3)(4x² + ax + 18)
Comparing coefficients of x²:
-30 = (4 x -3) + a
a = -18
4x³ - 30x² + 72x - 54 = (x - 3)(4x² - 18x + 18)
(x - 3)(4x² - 18x + 18) = 0

You will notice that (4x² - 18x + 18) has no solutions in the real number plane, so x = 3 is where your lone stationary point is located.

 

[Zak Ambrose]

i found 2 values for x
3/2 and 3.

 

[Chinmoku03]

Lol, I was retarded and forgot to use the quadratic formula XD

So yeah, use the quadratic formula on 4x² - 18x + 18. You will find that x = 3 and x = 3/2. So (x - 3)(4x² - 18x + 18) simplifies down to (x - 3)(x - 3)(x - (3/2)), which is (x - (3/2))(x - 3)². So you would then solve (x - (3/2))(x - 3)² = 0 to find the stationary points.

 

[P.T.F.E]

and im left with the answers +- 2, +- 5.
thank you very much.

what about x^3 + x^2 - 5x - 6 = 0 ?
id assume id have to factorize.

wait dont worry, end up with complex roots or sumin.

what about x^4 - 2x^3 = 0
i would factorise and get x^3(x-2) = 0
x = 0 or 2, i know the answers right, just checking if my working is ok

u probalbly dnt use this but i just had to post this for pple like me wanting to know the anwer it is 3 unit stuff but x^3 + x^2 - 5x - 6=0

factors of -6 where x^3 + x^2 - 5x - 6=0
when x = -2
-8 +4 + 10 - 6 =0
therefore (x+2) is a factor
remember (x-a) into P(a)=0
divided x+2 into x^3 + x^2 - 5x - 6=0

and u get (x+2)(x^2-x-3)
so that is that factorised!!

 

[addikaye03]

cool.
last question.
y = (x-1)(x-3)^3
sketch after finding dy/dx, x intercepts, y intercept, stationary points.
i found the x intercepts, 1 and 3 obviously.
y intercept is 27.

im just having trouble find the stationary points, another cubic question.

stationary points occur for
f'(x) = 0
using product rule
3(x-1)(x-3) + (x-3)^3 = 0
expanding i get
4x^3 - 30x^2 + 72x - 54 = 0
where do i go from here?

3(x-1)(x-3) + (x-3)^3 = 0------ Factorise (x-3) since common
therefore,(x-3)(3(x-1)+(x-3)^2)=0--------Expand what is inside the bracket, u'll get a non-factorable quadratic..
Skipping steps but discriminant will be be <0 therefore no roots
therefore x=3, sub this point into f(x) to find y- coordinate.