SOLUTION: I can't figure this one out, can you help me. Find the equation of the quadratic function with a vertex at V(1,3) passing thru the point P(2,1).Write answer in the form y=ax^2 +

Question 29577: I can't figure this one out, can you help me.
Find the equation of the quadratic function with a vertex at V(1,3) passing thru the point P(2,1).Write answer in the form y=ax^2 + bx + c.
Thanx for your help!
Answer by longjonsilver(2297) (Show Source): You can put this solution on YOUR website!
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... .

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 , we have:

(0,1):
--> 1 = c

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

(2,1):
--> 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

jon.

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