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Maths Methods 1/2
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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
vce-maths
vce-methods
Asked
Feb 13, 2011
by
Hayden
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270
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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. :)
Answered
Feb 16, 2011
by
Snorlax
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1,160
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## Maths Methods 1/2

Answered## 1 Answer