# VCE Maths questions help (1 Viewer)

#### InteGrand

##### Well-Known Member
Re: maths questions help

The graph of y = x^4 - 2x - 12 has 2 x-intercepts
a. construct a table of values for this polynomial rule for x = -3,-2,-1,0,1,2,3
b. Hence state an exact solution to the equation x^4 - 2x - 12 = 0
c. State an interval within which the other root of the equation lies and use the methods of bisection to obtain an estimate of this root correct to 1 decimal place

I get how to do part a and b, but i'm finding part c extremely challenging and confusing.
If you know a place where the polynomial is positive and another place where it is negative (and the root you already found does not lie between these two places), then the other root lies between these two numbers. You can then use the bisection method (check your textbook if you haven't learnt it yet).

#### boredsatan

##### Member
Re: maths questions help

I read my textbook but I still don't understand the bisection methods.

#### InteGrand

##### Well-Known Member
• kawaiipotato

#### boredsatan

##### Member
Re: maths questions help

The graph of y = x^4 - 2x - 12 has 2 x-intercepts
a. construct a table of values for this polynomial rule for x = -3,-2,-1,0,1,2,3
b. Hence state an exact solution to the equation x^4 - 2x - 12 = 0
c. State an interval within which the other root of the equation lies and use the methods of bisection to obtain an estimate of this root correct to 1 decimal place

I get how to do part a and b, but i'm finding part c extremely challenging and confusing.

#### InteGrand

##### Well-Known Member
Re: maths questions help

The graph of y = x^4 - 2x - 12 has 2 x-intercepts
a. construct a table of values for this polynomial rule for x = -3,-2,-1,0,1,2,3
b. Hence state an exact solution to the equation x^4 - 2x - 12 = 0
c. State an interval within which the other root of the equation lies and use the methods of bisection to obtain an estimate of this root correct to 1 decimal place

I get how to do part a and b, but i'm finding part c extremely challenging and confusing.
If you know a place where the polynomial is positive and another place where it is negative (and the root you already found does not lie between these two places), then the other root lies between these two numbers. You can then use the bisection method (check your textbook if you haven't learnt it yet).
.

#### boredsatan

##### Member
Re: maths questions help

Problem is i'm still not understanding (for whatever reason it's just not making sense to me) the bisection methods even after reading my textbook and the summary you linked me

#### pikachu975

##### I love trials
Moderator
Re: maths questions help

I'm going to assume bisection method is the same as halving the interval:

You have two points, one positive, one negative, e.g. (0,3) and (4, -4).
Then you bisect this interval and your new x value is x = (0+4)/2 = 2

For example if you sub in x = 2 you get y = 1.5, that means x = 2 is a closer approximation to the root than x = 0 because y = 1.5 is closer than y = 3 to the x axis (the root). So your new interval becomes 2 < x < 4

Hope this helped, don't know how else to explain it.

• InteGrand

#### boredsatan

##### Member
Re: maths questions help

How do you efficiently study for a maths assessment/test?

#### Drongoski

##### Well-Known Member
Re: maths questions help

How do you efficiently study for a maths assessment/test?
Very open-ended question. Not easy to provide a useful and meaningful answer.

#### boredsatan

##### Member
Re: maths questions help

Very open-ended question. Not easy to provide a useful and meaningful answer.
Does it really matter if I asked the question?

#### Drongoski

##### Well-Known Member
Re: maths questions help

Different people will provide different advice. Advice will depend on where you are at currently.

In general, make sure you are on top of the topics you are going to be examined on. Depending on how much time you have before the test, make sure you get on top of areas you may be weak in currently. Then do as many of the basic exercises (if you have not already done this) and then progress to more challenging questions after that - these harder questions can be from your books and preferably from your school's past, but more recent, test questions. Also don't stay up too late. When you know your stuff, you are more confident and less stressed and conversely. When you are less stressed, you tend to perform better.

Good luck.

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#### boredsatan

##### Member
Re: maths questions help

Different people will provide different advice. Advice will depend on where you are at currently.

In general, make sure you are on top of the topics you are going to be examined on. Depending on how much time you have before the test, make sure you get on top of areas you may be weak in currently. Then do as many of the basic exercises (if you have not already done this) and then progress to more challenging questions after that - these harder questions can be from your books and preferably from your school's past, but more recent, test questions. Also don't stay up too late. When you know your stuff, you are more confident and less stressed and conversely. When you are less stressed, you tend to perform better.

Good luck.
thanks #### boredsatan

##### Member
Re: maths questions help

What's the stationery point of inflection for y = 8(x+1)^3 - 1

#### boredofstudiesuser1

##### Active Member
Re: maths questions help

What's the stationery point of inflection for y = 8(x+1)^3 - 1
y' = 24(x+1)^2
When y'=0,
x = -1

y'' = 48(x+1)
When y''=0,
x = -1

So we can see x = -1 is both a stationary point and a point of inflexion (horizontal p.o.i.)

When x = -1, y = -1

Therefore, the stationary point of inflexion occurs at (-1,-1)

#### InteGrand

##### Well-Known Member
Re: maths questions help

y' = 24(x+1)^2
When y'=0,
x = -1

y'' = 48(x+1)
When y''=0,
x = -1

So we can see x = -1 is both a stationary point and a point of inflexion (horizontal p.o.i.)

When x = -1, y = -1

Therefore, the stationary point of inflexion occurs at (-1,-1)
Note that in general, y' = 0 and y" = 0 at some point is not sufficient for that point to be an inflection (for example, consider y = x^4 at x = 0).

In this example, the third derivative is non-zero at x = -1, which implies it is an inflection. Or you can note that the second derivative changes sign about x = -1.

#### boredofstudiesuser1

##### Active Member
Re: maths questions help

Note that in general, y' = 0 and y" = 0 at some point is not sufficient for that point to be an inflection (for example, consider y = x^4 at x = 0).

In this example, the third derivative is non-zero at x = -1, which implies it is an inflection. Or you can note that the second derivative changes sign about x = -1.
Yeah, sorry, forgot to put that step in.

#### boredsatan

##### Member
Re: maths questions help

y' = 24(x+1)^2
When y'=0,
x = -1

y'' = 48(x+1)
When y''=0,
x = -1

So we can see x = -1 is both a stationary point and a point of inflexion (horizontal p.o.i.)

When x = -1, y = -1

Therefore, the stationary point of inflexion occurs at (-1,-1)
Thanks for helping

#### boredsatan

##### Member
Re: maths questions help

1. The line with equation y = mx is tangent to the circle with centre (10,0) and radius 5 at the point P(x,y)
a. find the equation of the circle
b. show that the x-coordinate of the point P satisfies the equation (1+m^2)x^2 - 20x + 75=0
c. use the discriminant for this equation to find the exact value of m
d. find the coordinates of P (there are two such points)
e. find the distance of P from the origin

3. A circle has centre at the origin and radius 5. The point P(3,4) lies on the circle
a. Find the gradient of the line segment OP, where O is the origin
b. Find the gradient of the tangent to the circle at P
c. Find the equation of the tangent at P
d. If the tangent crosses the x-axis at A and the y-axis at B, find the length of line segment AB

5. An equilateral triangle ABC circumscribes the circle with equation x^2 + y^2 = a^2. The side BC of the triangle has equation x = -a
a. Find the equations of AB and AC
b. Find the equation of the circle circumscribing triangle ABC

7. For the curve with equation y = square root (x) - 1 and the straight line with equation y = kx, find the values of k such that:
a. the line meets the curve twice
b. the line meets the curve once

Thanks!!

#### boredsatan

##### Member
Re: maths questions help

1. The line with equation y = mx is tangent to the circle with centre (10,0) and radius 5 at the point P(x,y)
a. find the equation of the circle
b. show that the x-coordinate of the point P satisfies the equation (1+m^2)x^2 - 20x + 75=0
c. use the discriminant for this equation to find the exact value of m
d. find the coordinates of P (there are two such points)
e. find the distance of P from the origin

3. A circle has centre at the origin and radius 5. The point P(3,4) lies on the circle
a. Find the gradient of the line segment OP, where O is the origin
b. Find the gradient of the tangent to the circle at P
c. Find the equation of the tangent at P
d. If the tangent crosses the x-axis at A and the y-axis at B, find the length of line segment AB

5. An equilateral triangle ABC circumscribes the circle with equation x^2 + y^2 = a^2. The side BC of the triangle has equation x = -a
a. Find the equations of AB and AC
b. Find the equation of the circle circumscribing triangle ABC

7. For the curve with equation y = square root (x) - 1 and the straight line with equation y = kx, find the values of k such that:
a. the line meets the curve twice
b. the line meets the curve once

Thanks!!
Bump!!

#### Drongoski

##### Well-Known Member
Re: maths questions help

Q1. Eqn of circle: (x-10)^2 + y^2 = 5^2; where line meets this circle:
(x-10)^2 + (mx)^2 = 5^2 ==> (1+m^2)x^2 -20x + 75 = 0
(this is a quadratic equation with a repeated roots; discriminant = 0)

.: discriminant = (-20)^2 - 4 x (1+m^2) x 75 = 0 ==> m = +- (1/sqrt(3))

.: lines are: y = +- (1/sqrt(3)) x

.: the original quadratic equation becomes: (4/3)x^2 - 20x + 75 = 0 or 4x^2 - 60x + 225 = 0

i.e. (2x - 15)^2 = 0 ==> x = 15/2 ==> y = +- (1/sqrt(3)) times 15/2

.: you get co-ords of a couple of points, the points of contact of the tangents to the circle.

Sorry - I'll let others do the rest.

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