Homework Clinic
Mathematics Clinic => Grade 10 Mathematics => Topic started by: coco on Jun 18, 2013
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Please help me!
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x(x-4)=0 or x^2-4x=0
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If you are looking for a quadratic polynomial then first look at the standard equation
y = a*(x - h)^2 + k, where (h,k) is the vertex. So we need to find a. To do this, we use
the fact that the roots of the polynomial are 0 and 4, so we know that (0,0) and (4,0)
are on the graph. Now plug in one of these points, say (0,0), and (h,k) = (2, -8) into
the general equation:
0 = a*(0 - 2)^2 + (-8) ---> 0 = a*4 - 8 ---> 4a = 8 ---> a = 2.
Thus the equation of the quadratic polynomial is f(x) = 2*(x - 2)^2 - 8.
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Given that the roots at x = 0 and x = 4, x - 0 = x and x - 4 are factors of the quadratic equation that makes the required graph. So, we see that for some constant a the quadratic equation is:
f(x) = ax(x - 4).
Notice that this extra constant a is here because there are infinitely many quadratics with x = 0 and x = 4 as roots, only differing by a constant multiple of one another.
Since the quadratic has a vertex of (2, -8), the quadratic passes through (2, -8) (since all quadratics pass through their vertex) and f(2) = -8. Plugging in x = 2 into the above equation gives:
f(2) = a(2)(2 - 4) = -4a ==> -4a = -8 ==> a = 2.
Therefore, the required equation is:
f(x) = 2x(x - 4) = 2x^2 - 8x.
I hope this helps!