Question

Write an equation for a quadratic function that has x intercepts (-3, 0) and (5, 0)

Answer 1

One possible equation is
`f(x) = (x + 3)\, (x - 5)`, which is equivalent to
`f(x) = x^(2) - 2\, x - 15`.

Explanation:

The factor theorem states that if
`x = x_(0)` (where
`x_(0)` is a constant) is a root of a function,
`(x - x_(0))` would be a factor of that function.

The question states that
`(-3,\, 0)` and
`(5,\, 0)` are
`x`-intercepts of this function. In other words,
`x = -3` and
`x = 5` would both set the value of this quadratic function to
`0`. Thus,
`x = -3\!` and
`x = 5\!` would be two roots of this function.

By the factor theorem,
`(x - (-3))` and
`(x - 5)` would be two factors of this function.

Because the function in this question is quadratic,
`(x - (-3))` and
`(x - 5)` would be the only two factors of this function. In other words, for some constant
`a` (
`a \\e 0`):

`f(x) = a\, (x - (-3))\, (x - 5)`.

Simplify to obtain:

`f(x) = a\, (x + 3)\, (x - 5)`.

Expand this expression to obtain:

`f(x) = a\, (x^(2) - 2\, x - 15)`.

(Quadratic functions are polynomials of degree two. If this function has any factor other than
`(x - (-3))` and
`(x - 5)`, expanding the expression would give a polynomial of degree at least three- not quadratic.)

Every non-zero value of
`a` corresponds to a distinct quadratic function with
`x`-intercepts
`(-3,\, 0)` and
`(5,\, 0)`. For example, with
`a = 1`:

`f(x) = (x + 3)\, (x - 5)`, or equivalently,

`f(x) = x^(2) - 2\, x - 15`.