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||Définition d'une variable || Algèbre || Arithmetique || Logique ||La raison basée sur les règles et modelés||

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Read  logic chapters 1 to 5  in online volume Three Skills for Algebra  for greater skills & confidence in  work 
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Read notes and textbooks like a lawyer, so that no nuance, no subtlety and no clause escapes your attention. 

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 Logic chapters 1 to 5  re- appear not in sequence, as is or longer,  in  Volume 1A,  Pattern Based Reason, Bon Appetite.

Logic Mastery
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After logic  (a) continue reading Three Skills for Algebra, chapters 8 to 14  and do so alongside site area on solving liinear Equations ; or (b) see this calculus starter lesson and Volume 3, Why Slopes  & More Math, chapters 2 to 6;

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Discussion of the standard form y = a[(x-h)2+k]

Dependence on Parameters, Extrema and Zeros from standard form. Special Case of the Quadratic formula for the standard form.

The standard-form-for-graphing

y = a[(x-h)2+k]

represents two the result of two operations on the curve y = x2, a parabola that opens up with an axis of symmetry x = 0. The case a = 0 would give y = 0. So we suppose a is non-zero.

The first operation translates the points on y = x2. by adding (h,k) to them. That moves the axis of symmetry to x = h and raises or lowers the height of each point by  |k| = the magnitude of k.

The graph of  y = (x-h)2+k is parabola with minimum at x = h and an axis of symmetry x = h. This parabola and its axis of symmetry together look like an upward pointing pitchfork

If k > 0, the curve y = x2 is moved upward, and the equation

0 = y = (x-h)2+k 

has no solutions. So there are no x-intercepts.

If k = 0, the curve is not moved upward, and the equation

0 = y = (x-h)2+k = (x-h)2

has one solution, given by x - h = 0.  So there is only one x-intercept, namely x = h.

Finally, if k < 0, then the equation

0 = y = (x-h)2+k

has solutions, given by the values of x which satisfy

(x-h)2 = -k > 0

which are provided by the two equations  x - h = (-k)½  and x - k = (-k)½. Hence x = h plus or minus (-k)½ gives two solutions of

0 = y = (x-h)2+k

and hence the location of two x-intercepts. 

In all three cases, the graph of  y = (x-h)2+k is parabola with minimum at x = h and an axis of symmetry x = h. This parabola and its axis of symmetry together look like an upward pointing pitchfork.

The first operation translates the points on y = x2. by adding (h,k) to them. That moves the axis of symmetry to x = h and raises or lowers the height of each point by  |k| = the magnitude of k. The value of k determines whether or not the parabola part crosses the x-axes, twice, once or nonce times.

The second operation follows the translation by a vertical scaling of the y-values or coordinates, that is a vertical multiplication by the nonzero scale factor a of the points on the upward direct pitchfork  y = (x-h)2+k to obtain the graph of y = a[ (x-h)2+k.].

 The inclusion of the nonzero factor a  implies 0 = a[ (x-h)2+k.].when and only when  0 = [ (x-h)2+k]. So the previous analysis gives the number and/or  locations of the zeroes or y-intercepts. The number and/or locations depend only the values of h and k.

Now if a > 0, the second operation results in an parabola that opens up with a axis of symmetry x = h.  The case a > 1 implies the parabola height increases more rapidly than that of the parabola y = (x-h)2+k. The case 0 < a < 1 implies the parabola height increases less rapidly than that of the parabola y = (x-h)2+k

Now if a < 0, the second operation results is equivalent to multiply first by |a| and then multiplying by -1.   After both multiplications, the case |a| > 1 implies the parabola height decreases more rapidly than that of the parabola y = (x-h)2+k. The case 0 < |a| < 1 implies the parabola height decreases less rapidly than that of the parabola y = (x-h)2+k. The axes of symmetry together the graph of y = a[ (x-h)2+k.] looks like a pitchfork that points and opens downward.

Location of Zeroes

If k < 0, then y = (x-h)2 + k = 0 when and only when  (x-h)2 = -k or 
x-h =±   __
Ö-k
  or   x = h ±   __
Ö-k

This gives the first way to solve a[(x-h)2 + k ] = 0 or  ax2+bx+c = 0 when ax2+bx+c = a[(x-h)2 + k ]. The solutions are equidistant from the axis of symmetry, the line x = h.

 

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