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For the quadratic function y=ax^2+bx+c we know

the vertex is at (x,y)=(2,3)
there are two real roots which are integers one of which is x=1

What are the values of a,b,c

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1 Answer

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Are you familiar with the turning point from of a quadratic?

Quadratics can be rewritten in the form:

y = k(x - m)^2 + n, where m is the x-coordinate of the vertex, n is the y-coordinate of the vertex and k is known as a 'dilation factor'.

Since we know the vertex is at (2,3), then our equation becomes y = k(x - 2)^2 + 3

We are also told that one of the roots is x = 1, which basically means that x = 1 is one of the x-intercepts. Hence, we know that coordinate (1 , 0) belongs to this quadratic.

Substituting this into the new equation means that 0 = k(1 - 2)^2 + 3 and all we have to do is solve for k, which equals -3.

We now have: y = -3(x - 2)^2 + 3

A simple expansion yields: y = -3x^2 + 12x - 9

We do something called 'equating the coefficients' and we easily determine with the original equation that a = -3 , b = 12 and  c = -9.

Hopefully this reply wasn't too late. If it was, hopefully somebody else finds it useful. :)

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