# The graph of #y=ax^2+bx# has an extremum at #(1,-2)#. Find the values of a and b?

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From the given can substitute 1 for x and 2 for y and write the following equation:

Subtract equation [1] from equation [2]:

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To find the values of ( a ) and ( b ), we use the fact that the extremum of the parabola occurs at the vertex, which is given as (1, -2).

Given the general form of a quadratic function ( y = ax^2 + bx ), the x-coordinate of the vertex is ( -\frac{b}{2a} ).

From the given information, we know that the x-coordinate of the vertex is 1. So,

[ -\frac{b}{2a} = 1 ]

This implies ( b = -2a ).

Also, at the vertex, the y-coordinate is -2. So,

[ a(1)^2 + b(1) = -2 ] [ a + b = -2 ]

Substituting ( b = -2a ) into this equation:

[ a + (-2a) = -2 ] [ a - 2a = -2 ] [ -a = -2 ] [ a = 2 ]

Substitute ( a = 2 ) into ( b = -2a ):

[ b = -2(2) ] [ b = -4 ]

Therefore, the values of ( a ) and ( b ) are ( a = 2 ) and ( b = -4 ).

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