Question

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

x	f(x)
-8	-14
-7	-4
-6	2
-5	4
-4	2

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

Explanation:

Step1: Find the vertex of the parabola

A quadratic function is symmetric about its vertex. Looking at the table, the function values increase to \( x = -5 \) (where \( f(-5) = 4 \)) and then decrease, so the vertex is at \( (-5, 4) \). The vertex form of a quadratic function is \( f(x) = a(x - h)^2 + k \), where \( (h, k) \) is the vertex. So here, \( h = -5 \) and \( k = 4 \), so the equation becomes \( f(x) = a(x + 5)^2 + 4 \).

Step2: Find the value of \( a \)

We can use another point from the table to find \( a \). Let's use the point \( (-4, 2) \). Substitute \( x = -4 \) and \( f(x) = 2 \) into the equation:

$$ 2 = a(-4 + 5)^2 + 4 $$

$$ 2 = a(1)^2 + 4 $$

$$ 2 = a + 4 $$

Subtract 4 from both sides: \( a = 2 - 4 = -2 \).

Step3: Write the final equation

Substitute \( a = -2 \), \( h = -5 \), and \( k = 4 \) into the vertex form:

$$ f(x) = -2(x + 5)^2 + 4 $$

We can expand this to standard form to check:

$$ f(x) = -2(x^2 + 10x + 25) + 4 = -2x^2 - 20x - 50 + 4 = -2x^2 - 20x - 46 $$

Let's verify with another point, say \( x = -6 \): \( f(-6) = -2(-6 + 5)^2 + 4 = -2(1) + 4 = 2 \), which matches the table.

Answer:

\( f(x) = -2(x + 5)^2 + 4 \) (or expanded form \( -2x^2 - 20x - 46 \))