# Here is a graph before me. The points on the graph are: (4,-4) (3,-3) (5,-3) (2,0) (6,0) (1,5) (7,5). One line runs through all the points making a v shape. the question says "Given the graph...

Here is a graph before me. The points on the graph are: (4,-4) (3,-3) (5,-3) (2,0) (6,0) (1,5) (7,5). One line runs through all the points making a v shape. the question says "Given the graph below, write the functon for f(x) in standard form."

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Given the points (1,5),(2,0),(3,-3),(4,-4),(5,-3),(6,0),(7,5) write the function for the curve that contains these points.

This is a parabola. You can verify this using the method of finite differences.

The vertex is at (4,-4), it opens up so the leading coefficient is positive, and the zeros are at x=2 and x=6.

One way to find the equation is to put in vertex form `y=a(x-h)^2+k` where the vertex is at (h,k). So

`y=a(x-4)^2-4` Now we can put in a known (x,y) pair to solve for a: Choosing (2,0) we have

`0=a(2-4)^2-4` or 0=4a-4 ==> a=1

The equation is vertex form is `y=(x-4)^2+4` . Writing in standard form by expanding the binomial we get:

`y=x^2-8x+12`

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Alternatively, since we know the zeros we can write in factored form `y=a(x-p)(x-q)` where p,q are the zeros. Here we have:

y=a(x-2)(x-6) Substituting a known (x,y) pair, say (4,-4) to solve for a we get:

-4=a(4-2)(4-6) or -4=-4a ==> a=1 so in factored form we have

y=(x-2)(x-6) Multiplying the right hand side to get standard form we get `y=x^2-8x+12` as before.

The graph: