A quadratic function has zeros at 1 and -3 and passes through the point (2,10).

(1) First we can write the function in intercept form: `y=a(x-p)(x-q)` where `p,q` are the x-intercepts (zeros, roots, etc...)

Thus we have `y=a(x-1)(x+3)` (Note that if k is a zero of a function, then (x-k)...

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A quadratic function has zeros at 1 and -3 and passes through the point (2,10).

(1) First we can write the function in intercept form: `y=a(x-p)(x-q)` where `p,q` are the x-intercepts (zeros, roots, etc...)

Thus we have `y=a(x-1)(x+3)` (Note that if k is a zero of a function, then (x-k) is a factor)

We need to solve for the coefficient a; we use the point (2,10):

`10=a(2-1)(2+3)`

`10=5a`

`a=2`

Thus we have `y=2(x-1)(x+3)`

(2) From here we can write in general form: `y=ax^2+bx+c` by multiplying out the factors:

`y=2(x-1)(x+3)`

`y=2[x^2+2x-3]`

`y=2x^2+4x-6`

(3) Now we can write in vertex form `y=a(x-h)^2-k` by completing the square:

`y=2[x^2+2x-3]`

`y=2[x^2+2x+1-1-3]`

`y=2[(x+1)^2-4]`

`y=2(x+1)^2-8`

**Thus the desired function is `y=2(x+1)^2-8` **

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We could also have found the general form by using the three points (1,0),(-3,0) and (2,10). Plug each into the general form `y=ax^2+bx+c` . You will get three equations with the three unknowns a,b, and c. Solve these equations simultaneously (using linear combinations, substitution, or matrices)

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**Quickest way**

You could get directly to the vertex form: knowing the two zeros are 1 and -3, the axis of symmetry must lie midway between them at x=-1. Then using the points (1,0) and (2,10) and the vertex form you get:

`0=a(1+1)^2+k`

`10=a(2+1)^2+k`

Solving these simultaneously leads to a=2,k=-8 and the result.