C1 C2 M1 Revision

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Core 1

Linear Graphs and Equations

For any straight line, the gradient (M) is: dy/dx or difference in y/difference in x which is (y2-y1)/(x2-x1) Equation of a line: y=mx+c which is used when the gradient and intercept is known or y-y1=m(x-x1) when the gradient and the co-ordinates (x1,y1) of a single point that the line passes through is known. You'll need to learn this equation. [The equation of the line can be kept in this form unless stated in the exam. (reduces error chance) Also, usually there is a working mark, so state the fact that the gradient is difference in y/difference in x] The mid-point of two graphs is found by (x1+x2)/2 , (y1+y2)/2 in the form (x,y) Lines with the same gradient are parallel, while lines with gradients that are negative reciprocals of each other is perpendicular to it. (perpendicular means at a right angle to) [they usually want you to state that 'the perpendicular line's gradient is the negative reciprocal', so stick it in. It is also usefull to draw out diagrams if they ask about right angle triangles, (usually something to do with negative reciprocals rather than pythagoras.)] In 2D Lines that are not parallel must intesect, the point of intersection can be found by simultaneous equations by: • equating coefficients of the two lines

• substituing one equationinto the other
• equating both equations for y.


surd form is exact. they involve irrational roots, which are roots that cannot be expressed as fractions as they are irrational for example: √5' (for clarification purposes ' marks the end of the root) • √a' x √b' = √a x b' = √ab'

• √a' / √b' = √a/b'
• √a+b' ≠ √a'+√b'
It is often more useful when denominator of a fraction is rationalised. This is done by multiplying the top and bottom by the conjugate, as the product of two conjugates is always rationalised because (a+b)(a-b)=(a^2)-(b^2) and a surd^2 is always rational. (^2 means squared)

Quadratic Graphs and Equations (and Further Equations)

The graph of a quadratic is a parabola, they can be factorised into two linear factors. Parallel beams of light on a parabola reflector reflect through the same point. Any moving object in gravity follow a parabolic curve. A plane that cuts a cone makes a parabola shape parallel to the cone. A quadratic can be solved by:

• factorising (finding y=(x+r1)(x+r2) where r=root)
• completing the square (in the form y=k(x-p)^2 +q note the -ve infront of the p ) • using the quadatic equation. (needs to be learnt)
the graph y=ax^2 +bx+c lets you find the y-intercept, c.
quadratics in completed square form have a translation of [p over q] (dont know how to format) with the vertex at (p,q) symmetry at x=p [remember the word translation, question on it every exam without fail.] quadratics completed in factorised form lets you find the root(s)/solution(s)/x-intercept(s)/when y is = 0 The discriminant b^2-4ac can be used to find the number of roots: • positive answer = 2 real roots

• 0 = one root / repeated roots.
• negative answer = no 'real' roots. (you cant square a negative number, so quadatic formula wouldnt work ∴ no real roots) [if the exam doesnt say how many roots, just that they're real its ≥0, they also usually want a statement stating the bullet points above.] If a pair of simultaneous equations lead to a quadratic, then the discriminant shows the relationship between the graphs. The same rules about the discriminant applies, ie. number of intersections.


finding solution set of an inequality means to solve an inequality. linear equality = both expressions are linear (GCSE stuff)
remember that multiplying/dividing by negative numbers means that the inequality symbol is reversed. The discriminant can also be turned into an inequality. > 0, = 0, < 0, ≥0 The solution can be represented on a number line, graphically or using sign diagrams showing critical values for both linear and quadratic...
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