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

User Slurry
by
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1 Answer

10 votes

Answer:

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.

User Mkafiyan
by
6.6k points
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