The table below shows selected values for a polynomial function f of degree 3. | x | f(x) … — SAT

The table below shows selected values for a polynomial function $f$ of degree 3.

$x$	$f(x)$
$-2$	$0$
$1$	$0$
$3$	$0$
$4$	$18$

What is the value of $f(0)$ ?

Cevap: 6

Cevap

The value of

f(0)

is 6.

The table shows the roots of the cubic function

f(x)

are

x = -2

x = 1

, and

x = 3

. This allows the function to be written in factored form as

f(x) = a(x + 2)(x - 1)(x - 3)

. Using the point

(4, 18)

from the table, we solve for

a

18 = a(4 + 2)(4 - 1)(4 - 3) \Rightarrow 18 = 18a \Rightarrow a = 1

. The function is

f(x) = (x + 2)(x - 1)(x - 3)

. Substituting

x = 0

gives

f(0) = (2)(-1)(-3) = 6

Adım Adım Çözüm

Identify the roots of the polynomial from the table.

The roots of the function are

x = -2

x = 1

, and

x = 3

The table shows that

f(x) = 0

at these values, indicating they are the

x

-intercepts or roots.

Write the general factored form of a degree 3 polynomial with these roots.

f(x) = a(x + 2)(x - 1)(x - 3)

A cubic polynomial with roots

r_1

r_2

, and

r_3

can be expressed as

f(x) = a(x - r_1)(x - r_2)(x - r_3)

for a constant coefficient

a

Find the value of the constant coefficient

a

using a known point from the table.

a = 1

Substituting the table values

x = 4

and

f(4) = 18

yields

18 = a(4 + 2)(4 - 1)(4 - 3)

, which simplifies to

18 = 18a

, so

a = 1

Calculate the value of

f(0)

using the fully defined function.

f(0) = 6

Substituting

x = 0

into the equation

f(x) = (x + 2)(x - 1)(x - 3)

gives

f(0) = (0 + 2)(0 - 1)(0 - 3) = 2 \times (-1) \times (-3) = 6

Anahtar Kavram

Using polynomial roots and extra points to define a polynomial function and evaluate it.

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