## Sketching derivative graphs (cubics)

The important thing to remember here is that we are doing a sketch. It does not have to be accurate, but id does have to conatian all the key points.

Hints:

- A cubic differentiates to a quadratic (parabola)
- A the turning points have a gradient of zero and will move to the x axis on the derivative sketch
- A point with a positive gradient will be above the x axis in the sketch
- A point with a pnegative gradient will be below the x axis in the sketch

Examples

Describe and sketch the derivative graph for the function below

The critical points are , because this is where the turning points occur.

These points will be on the x axis in the sketch.

For the gradient is positive.

These points will be above the x axis in the sketch.

For the gradient is negative .

These points will be below the x axis in the sketch.

The graph is a cubic so the sketch will be a parabola

### Practice

## Simultaneous Equations: Substitution Method

This is exactly the same method we use to find where two lines intersect.

Process:

- Rearrange one of the equations to make one of the variables a subject
- Substitute for this variable into the other equation
- Solve the resulting equation
- Substitute these solutions into one of the original equations to find the values of the other variable

Example

Solve the system of equations

Substitute eqn 1 into eqn 2

Substituting into eqn 1 gives

Solve the system of equations

Rearrange eqn 2 to make y the subject

Substitute eqn 3 into eqn 1

Factorising gives

Substituting these values into eqn 3 gives

So the solutions are

Solve the system of equations

Rearranging eqn 2 to make y the subject gives

Substituting eqn 3 into eqn 1 gives

Substituting the value of x into eqn 3 gives

## Simultaneous equations: Elimination method

Simultaneous equations are sytems of equations where there are a number of variables.

Our task is to find the values of the variables that satisfy all the equations.

For this method we will solve equations of the type

With these equations we can do one of three things

- Multiply by a scaler.
- Add the equations together
- Subtract one equation from another

Multiplying by a scaler

eqn 1 multiplied by 6 gives

Adding two equations

eqn 1 + eqn 2 gives

Subtract one equation from another

eqn 1 – eqn 2 gives

### Type 1 Simultaneous equations

With this type of problem the y terms are the same size but different signs.

To solve these we:

- Add the equations together (this will cancel out the y terms)
- Solve the resulting equation to find the value of x
- Substitute the x value into one of the original equations
- Solve the resulting equation to find the value of y
- Check by substituting x and y into the other equation

Examples

Solve

This is a type 1 problem because we have two 4y’s, but one is positive and the other is negative.

So eqn 1 + eqn 2 gives

Substitute into eqn 1

gives

Check in eqn 2

true

Solve

This is a type 1 problem because we have two 5y’s, but one is positive and the other is negative.

So eqn 1 + eqn 2 gives

Substitute into eqn 2 (it is easier to deal with positives)

gives

Check in eqn 1

true

### Type 2 Simultaneous equations

With this type of problem the y terms are the same size and sign.

To solve these we:

- Subtract one equation from the other (this will cancel out the y terms)
- Solve the resulting equation to find the value of x
- Substitute the x value into one of the original equations
- Solve the resulting equation to find the value of y
- Check by substituting x and y into the other equation

Examples

Solve

This is a type 2 problem because we have two +2y’s.

So eqn 2 – eqn 1 gives

Substitute into eqn 1

gives

Check in eqn 2

true

Solve

This is a type 2 problem because we have two -4y’s.

So eqn 2 – eqn 1 gives

Substitute into eqn 1

gives

Check in eqn 2

true

### Type 3 Simultaneous equations

With this type of problem the y terms are different.

To solve these we:

- Multiply each equation by the other equations positive y coefficient
- Solve the resulting type 1 or type 2 problem

Examples

Solve

Multiply eqn 1 by 2 (that is the number of y’s in eqn 2)

Multiply eqn 2 by 4 (that is the number of y’s in eqn 1)

Equations 3 and 4 now make a type 2 problem

Subtracting eqn 3 from eqn 4 gives

Substitute into eqn 2

Check in eqn 1

true

Solve

Multiply eqn 1 by 3 (that is the number of y’s in eqn 2)

Multiply eqn 2 by 2 (that is the number of y’s in eqn 1)

Equations 3 and 4 now make a type 1 problem

Adding eqn 3 and eqn 4 gives

Substitute into eqn 2

Check in eqn 1

true

## Circle Theorems 2

Tangents to circles

#### Investigation

### Theorem 1

The length of the tangents from a point to a circle are of equal length

### Theorem 2

The Angle between a tangent and a radius is a right-angle

### Theorem 3

The angel between a tangent and a cord is equal to the angle subtended by the cord

## Examples

Given that PA is a tangent find the value ofSince PA is a tangent

Given that PA and PB are tangents find the value ofSince PA and PB are tangents they must both be of equal length. Therefore is isosceles.base angles in an isosceles triangle are equal.the angle between a tangent and a radius is a right-angle.

Find the value ofSince

#### Questions

## Trigonometry: Right-angled triangles Quesitons

For each set of questions find the value of

Make sure to write down all the working. It is a good habit to be rigorous in your answers.

Set 1

Set 2

Set 3

Set 4

Set 5

## Trigonometry and Right-angled triangles

#### Conceptual Understanding

If we find the ratio between a pair of sides on a triangle the value will be a constant even on larger triangles provided the triangles are similar to each other.

### The right-angled trigonometry triangle

The hypotenuse is the longest side and is opposite the right-angle

The opposite is the side that is opposite the given angle. (Or the angle we wish to find)

The adjacent is the side next to the given angle. (It is between the given angle and the right angle)

### The ratios SohCahToa

When solving a trigonometry right-angled triangle problem we need two pieces of information and then use the relevant ratio to find the third.

There are three possible scenarios.

- We are finding an angle
- We are finding a side that is a numerator in the ratio
- We are finding a side that is a denominator in the ratio

## Examples

Type 1 (The unknown is the numerator)

We have the hypotenuse and we want the opposite side. The only ratio with hypotenuse and opposite in it is the sine ratio.

Let’s fill out the equation

Type 2 (The unknown is the denominator)

We have the opposite and we want the adjacent side. The only ratio with adjacent and opposite in it is the tangent ratio.

Let’s fill out the equation

Type 3 (The angle is unknown)

We have the hypotenuse and adjacent side. The only ratio with adjacent and hypotenuse in it is the cosine ratio.

Let’s fill out the equation