Question 1986: Sorry, long time since I've done maths! The question is "find a formula for the quadratic function as shown." The intercepts are (x= -1,y=2,) ( x=1, y=0 )and (x= -3, y=0.) Thanks for your time and help.
Answer by longjonsilver(2297)   (Show Source):
You can put this solution on YOUR website! rightio..i understand now :-)
however, the point (-1, 2) is not an intercept, it is the maximum (or minimum) point on the curve...
ok. where an equation/line/curve crosses the x-axis is where y=0. Finding such values (called the roots of the equation) is the main aim in algebra like this.
typically we have a quadratic equation equal to zero which we factorise because then we have 2 parts that multiply to equal zero. For this to be true, one of those 2 parts must be zero itself. Which one? either.
so here, we work backwards...we know that the roots are x=1 and x=-3, so we know that
(x-1)(x+3) = 0
so, now we can multiply this out to give  . Collecting the 2 middle x-terms gives  -eqn1
so what is the equation? well it is  , but hold on...what about moving all the terms in eqn1 to the other side so we had  ? Here we have the equation 
So we have 2 quadratics that both go through the 2 roots mentioned. Here they are sketched...
As you can see, they are both symmetric about the max/min axis (at x=-1). So we need this third point to tell us which of the 2 curves, and hence, which of the 2 equations we had.
Note:
a u-shape quadratic has +ve x-squared term
a n-shape quadratic has -ve x-squared term
so, the point (-1, 2) should lie on the n-shaped curve and hence our equation was . However, looking at the curve, the point is more like (-1, 4). Oh well!
Hope this was of help to you.
cheers
Jon.
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