Question

there is a function f that is quadratic and some of its ordered pairs are shown in the table below.

x	f(x)
1	3
2	-6
3	-9
4	-6

write an equation for the function.
f(x)=\square

Explanation:

Step1: Identify vertex form

A quadratic function in vertex form is \( f(x) = a(x - h)^2 + k \), where \((h, k)\) is the vertex. From the table, the function is symmetric around \( x = 3 \) (since \( f(2) = f(4) = -6 \), \( f(1) = f(5) \) would be 3, etc.), so the vertex is \((3, -9)\). Thus, \( h = 3 \), \( k = -9 \), so \( f(x) = a(x - 3)^2 - 9 \).

Step2: Find 'a' using a point

Use the point \((0, 18)\) (when \( x = 0 \), \( f(x) = 18 \)). Substitute into the equation:
\( 18 = a(0 - 3)^2 - 9 \)
\( 18 = 9a - 9 \)
Add 9 to both sides: \( 27 = 9a \)
Divide by 9: \( a = 3 \).

Step3: Write the equation

Substitute \( a = 3 \), \( h = 3 \), \( k = -9 \) into vertex form:
\( f(x) = 3(x - 3)^2 - 9 \). Expand to standard form (optional, but let's check):
\( f(x) = 3(x^2 - 6x + 9) - 9 = 3x^2 - 18x + 27 - 9 = 3x^2 - 18x + 18 \). Verify with \( x = 1 \): \( 3(1) - 18(1) + 18 = 3 - 18 + 18 = 3 \), which matches.

Answer:

\( 3(x - 3)^2 - 9 \) (or \( 3x^2 - 18x + 18 \))