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Unformatted text preview: Quadratic equations and quadratic functions Introduction . A quadratic function is a function f of the form = ( ) f x + + a x 2 b x c (i), where a 0. A quadratic equation is an equation of the form = + + a x 2 b x c 0  (ii), where a 0. The real number solutions of equation (ii), if any, provide the x intercepts of the graph of the function f. Any real number x in the domain of a general function g, which has real number values, such that = ( ) g x 0 is called a zero of g. Hence any solution of the quadratic equation (ii) will be a zero of the quadratic function f defined by equation (i). Example 1 : Solve the quadratic equation =   2 x 2 x 3 0, and illustrate the solutions as the x intercepts of the graph of = y  2 x 2 x 3. Solution : By factoring the left side of the equation =   2 x 2 x 3 0, we see that it is equivalent to = ( ) 2 x 3 ( ) + x 1 0. This gives = x 3 2 or = x 1. _______ To sketch the graph of = y  2 x 2 x 3 we make use of the technique of completing the square . Given a pair of terms + x 2 2 p , we can add p 2 , which is "half the coefficient of x squared" to form the perfect square = + + x 2 2 p p 2 ( ) + x p 2 . = y  2 x 2 x 3 is equivalent to = y 2  x 2 x 2 3  (iii). Adding =  1 4 2 1 16 inside the bracket  x 2 x 4 will form the perfect square =  + x 2 x 2 1 16  x 1 4 2 . It is useful to incorporate this perfect square in a rearrangement of the right side of equation (iii), but we must make an adjustment to the constant  3 in the right side of (iii). By adding 1 16 inside the bracket we have effectively added = 2 1 16 1 8 to the right side of equation (iii) because of the 2 factor to the left of the bracket. Thus we shall have an equation equivalent to (iii) if we also subtract 1 8 "outside the bracket" to give = y  2  + x 2 x 2 1 16 1 8 3. Alternatively, introduce +  1 16 1 16 inside the bracket so that equation (iii) becomes: = y 2  +  x 2 x 2 1 16 1 16 3 , which gives: = y  2  + x 2 x 2 1 16 1 8 3. This equation can be written in the form = y 2  x 1 4 2 25 8 (iv). Using ideas about transformations of graphs of functions, the equation (iv) shows that the graph of = y  2 x 2 x 3 has the same shape as the parbabola = y 2 x 2 , because it can be obtained by first translating the curve = y 2 x 2 to the right by 1 4 unit to give the curve = y 2  x 1 4 2 , and then shifted = 25 8 3 1 8 units down to form the curve = y 2  x 1 4 2 25 8 . In particular, the lowest point or vertex of the parabola = y 2 x 2 moves from the origin to the point , 1 4 25 8 ....
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This note was uploaded on 11/12/2011 for the course MATH 111 taught by Professor Uri during the Spring '08 term at Rutgers.
 Spring '08
 Uri
 Equations

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