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Mathematics 7 Online
OpenStudy (anonymous):

h

OpenStudy (anonymous):

Start with y=ax^2 + bx + c

OpenStudy (campbell_st):

well the easy way to do this is to set up the equation using the x intercepts \[y = a(x + 3)(x -1)\] where a is a constant distributing gives \[y = a(x^2 + 2x -3) \] now substitute the values from the y intercept, x = 0 and y = 2 to find the value of the constant a. then just distribute and you'll have a quadratic meeting the given conditions.

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