Find the function y=ax^2 + bx + c that when graphed has an x-intercept of 1, a y-intercept of -2, and at the y-intercept the slope of the tangent is -1.

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We have to find the function y = ax^2 + bx + c such that when the graph of the function is drawn, it has an x-intercept of 1, a y-intercept of -2 and a tangent line with a slope of -1 at the y-intercept.

The x-intercept is 1

=> a(1)^2 + b(1) + c = 0

=> a + b + c = 0 ...(1)

The y-intercept is -2

=> -2 = a*0 + b*0 + c

=> c = -2 ...(2)

The slope of the tangent to the graph at x = c is given by f'(c)

f'(x) = 2ax + b

f'(0) = 2a*0 + b = -1

=> b = -1

Substitute b = -1 and c = -2 in (1)

=> a = 3

The equation of the function is y = 3x^2 - x - 2

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