When does a parabolic graph have integral x-intercepts and vertex?Given a quadratic function f(x)=ax2+bx+c for integers a and b and c, what must be true of a and b and c to guarantee the roots of f and the location and value of the minimum/maximum of f are integers? In terms of the graph of f, how can we guarantee the x-intercepts and the vertex will have integer coordinates?

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Porter Camacho · Accepted Answer

If m and n are two integers with the same parity, and k is a non-zero integer, thenf(x)=kx2−k(m+n)x+kmnhas integral roots namely m and n and the location of vertex namely m+n2 is also an integer.It remains to show that f(m+n2) is also an integer.Note thatf(m+n2)=k(m+n2)2+k(m+n)22+kmnis also an integer because m and n have the same parity.

Rowan Waller · Accepted Answer

Writing outax2+bx+c;;=;;a(x+b2a)2−b2−4ac4awe see the vertex of the graph of f must be(,,−b2a,−b2−4ac4a,,),which will be integral if 2a|b. Then the roots of f are given by−b2a±b2−4ac4a2which will give us integral x-intercepts if a|c and if b2−4ac is a square. Maybe there&#039;s a neater way to state last last requirement accounting for requirements that 2a|b and a|c.

When does a parabolic graph have integral x-intercepts

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