C1: Chapters 1-4 Revision

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C1: Chapters 1-4 Revision. Dr J Frost (jfrost@tiffin.kingston.sch.uk) . Last modified: 10 th October 2013. Solving simultaneous equations. Remember that the strategy is to substitute the linear equation into the quadratic one, then solve. ?. Expanding out correctly!. ?. Inequalities.

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### C1: Chapters 1-4 Revision

Dr J Frost (jfrost@tiffin.kingston.sch.uk)

Solving simultaneous equations

Remember that the strategy is to substitute the linear equation into the quadratic one, then solve.

?

Inequalities

• For inequalities in general:
• Multiplying/dividing both sides by a negative number flips the inequality.
• Don’t mix up AND and OR. “” is different from “”.

Always start by putting in the form or .If you have , ABSOLUTELY DON’T divide by , but write

Then factorise.

Then sketch.Your answer will either be , or “or”. Be sure to use the word ‘or’ in the latter one, since ‘and’ would be wrong.

Find the set of values of x for which

(a) 4x – 3 > 7 – x (b) 2x2 – 5x – 12 < 0 (c) both 4x – 3 > 7 – x and 2x2 – 5x – 12 < 0

?

?

?

Discriminant

• Whenever you see the words “equal roots”, “distinct/different roots” or “no roots”, you know you’ve got to calculate the discriminant, which is .
• It helps to explicitly write out your , and first before substituting into the discriminant.
• Be VERY careful with double (or even triple!) negatives.The discriminant of is .The discriminant of is 4.
• When you have ‘different roots’ or ‘no roots’, you’ll have a quadratic inequality. Solve in the same way as before. But remember your sketch is in terms of , not in terms of the original variable . So don’t be upset if your sketch has roots, even if the original question asks where your equation has no roots.

?

?

The equation , where k is a constant, has 2 different real solutions for x.

(a) Show that k satisfies

(b) Hence find the set of possible values of k.

?

• For cubics, think whether the / term is positive or negative. Cubics with positive will go uphill, and downhill otherwise.
• If , without fully expanding you can tell you’ll have a term, thus it goes downhill. Be careful though: in , the term will be positive!
• You can get the roots/-intercepts by setting to be 0. Imagine each factor/brackets being 0. So if , then the roots are
• For both quadratics and cubics, the curve touches the x-axis for a root if the factor is squared, and crosses if not repeated.
• Don’t forget the y-intercept!YOU WILL LOSE MARK(S) OTHERWISE.
• It’s quite acceptable to have algebraic expressions as roots/y-intercepts. The y-intercept is ? No problem!
• Don’t forget what a sketch of or looks like.

Sketching cubics

Sketch the following, ensuring you indicate the values where the line intercepts the axes.

y = (3-x)3

1

y = (x+2)(x-1)(x-3)

5

y = x(x+1)2

9

?

27

?

?

6

3

-1

-2

1

3

2

y = x(x-1)(2-x)

6

y = x(1 – x)2

10

y = (x+2)2(x-1)

?

?

?

1

-2

1

2

1

-4

7

y = -x3

11

y = (2-x)(x+3)2

3

y = x(2x – 1)(x + 3)

?

?

?

18

-3

2

0.5

3

4

y = x2(x + 1)

8

y = (x+2)3

12

y = (1 – x)2(3 – x)

?

?

?

8

3

-2

-1

1

3

Transforming Existing Graphs

a f(bx + c) + d

Bro Tip: To get the order of transformations correct inside the f(..), think what you’d need to do to get from (bx + c) back to x.

Step 1:

?

 c

Step 3:

?

↕ a

Step 4:

↑ d

?

Step 2:

?

↔ b

Transforming Existing Graphs

Here is the graph y = f(x). Draw the following graphs, ensuring you indicate where the graph crosses the coordinate axis, minimum/maximum points, and the equations of any asymptotes.

y

Bro Tip: Don’t get to transform the asymptotes! This horizontal asymptote won’t be affected by any transformations, but will by ones.

y = f(x)

(2, 3)

1

x

y = 2f(x+2)

y = -1

?

y

y

y

y = 0

y = -f(-x) – 1

x

6

1

?

y = f(2x)

-2

?

x

x

(1, 3)

y = -1

y = -2

(-2, -4)

Sketching Graphs by Considering the Transform

It’s often helpful to consider a simpler graph first, e.g. or , and then consider what transform we’ve done.

Sketch

?

Then clearly we’ve replaced with and added 4 to the result.

i.e.

?

Sketch y = x2 + 2x + 1

Sketch y = x2 + x – 2

y

?

y

?

1

x

x

-2

1

1

-2

Sketch y = -x2 + 2x + 3

Sketch y = 2x2 – 5x – 3

y

y

?

?

3

x

x

-1

3

-0.5

3

-3

Some quadratics have no roots. In which case, you’ll have to complete the square in order to sketch them. This tells you the minimum/maximum point.

e.g.

So minimum point is .

-intercept is 3

?

?

?

Sketch y = -x2 + 2x – 3

Sketch y = x2 – 4x + 5

y

?

y

?

5

(1,-2)

(2, 1)

-3

x