Question

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

Answer 1

Answer:

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

Step-by-step 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 $a$ ($a \ne 0$):

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

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,

$f(x) = x^{2} - 2\, x - 15$.