Question 29577
There are a couple of ways to do this. I shall show you the more visual method, so you can appreciate what quadratics are a bit more.


First, I shall show you the graph of the answer... {{{graph(300,300,-2,4,-4,4,-2x^2+4x+1, 1) }}}.


How do i know the graph is n-shaped rather than u-shaped? Well, plot the 2 points you know. If the vertex is at (1,3), there is no way that a u-shaped curve can pass through point (2,1) as well.


I have drawn a line at y=1, since you are given the point (2,1) is on the curve. Now, a quadratic is ALWAYS symmetric about the vertex. The vertex is at x=1: a given in the question. We are also given the point on the curve at x=2, so we instantly know that the point one less than the vertex (x=0) also has the y-value of 1... points (0,1) and (2,1) are symmetric.


So we have 3 points. We have an equation with 3 unknowns too... a, b and c. So we can now solve.


So, starting with {{{y = ax^2 + bx + c }}}, we have:


(0,1): {{{ 1 = a(0)^2 + b(0) + c }}}
 --> 1 = c


(1,3): {{{ 3 = a(1)^2 + b(1) + c }}}
 --> 3 = a + b + c
 --> 3 = a + b + 1
 --> a+b = 2 -->eqn1


(2,1): {{{ 1 = a(2)^2 + b(2) + c }}}
 --> 1 = 4a + 2b + c
 --> 1 = 4a + 2b + 1
 --> 4a+2b = 0
 --> 2a+b = 0 -->eqn2


Subtract eqn1 from eqn2 to give a = -2


Hence, using 2a+b=0 we get that 2(-2)+b = 0
-4 + b = 0
--> b = 4


So the formula is {{{ y=-2x^2 + 4x + 1 }}}


jon.