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Using transformation sketch the graph of each of the following. g ( x ) = - x 2 Solution (a) Depending on the placement of
(9x-3)(x+6)
Solve 2 x 10 - x 5 - 4 = 0 . Solution We can reduce this to quadratic in form using the substitution, u = x 5 u 2 = x 10 By using this substitution the equa
#7.
f(2)=3 and g(x)=x^2+1 then gof(2)
Properties of f( x ) = b x 1. The graph of f( x ) will always have the point (0,1). Or put another way, f(0) = 1 in spite of of the value of b. 2. For every possible b b x
Solve 2x2 = 128. (Points : 1) {±8} {±6} {±64} {±32}
Example Solve out the following system of equations. x 2 + y 2 = 10 2 x + y = 1 Solution In linear systems we had the alternative of using either method on any gi
(b+a)+[-(a+0+b)]=0
These are the only possibilities for solving quadratic equations in standard form. However Note that if we begin with rational expression in the equation we might get different so
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