Question

question
the function $f(x)$ is a quadratic function and a limited table of values is provided below. write the equation of the quadratic polynomial in standard form.

$x$	$f(x)$
$-2$	$3$
$-1$	$0$
$0$	$-1$
$1$	$0$
$2$	$3$
$3$	$8$

answer
attempt 1 out of 2
$f(x) = $

Explanation:

Step1: Recall quadratic standard form

A quadratic function in standard form is \( f(x) = ax^2 + bx + c \). We can use the table values to set up equations. Let's use the points \((-1, 0)\), \((0, -1)\), and \((1, 0)\) (or others, but these have simple \(x\) values).

Step2: Use \(x = 0\) to find \(c\)

When \(x = 0\), \(f(0) = a(0)^2 + b(0) + c = c\). From the table, \(f(0) = -1\), so \(c = -1\). Now the function is \(f(x) = ax^2 + bx - 1\).

Step3: Use \(x = 1\) to set up equation

When \(x = 1\), \(f(1) = a(1)^2 + b(1) - 1 = a + b - 1\). From the table, \(f(1) = 0\), so \(a + b - 1 = 0 \implies a + b = 1\) (Equation 1).

Step4: Use \(x = -1\) to set up equation

When \(x = -1\), \(f(-1) = a(-1)^2 + b(-1) - 1 = a - b - 1\). From the table, \(f(-1) = 0\), so \(a - b - 1 = 0 \implies a - b = 1\) (Equation 2).

Step5: Solve the system of equations

Add Equation 1 and Equation 2: \((a + b) + (a - b) = 1 + 1 \implies 2a = 2 \implies a = 1\). Substitute \(a = 1\) into Equation 1: \(1 + b = 1 \implies b = 0\).

Step6: Write the quadratic function

Now that \(a = 1\), \(b = 0\), and \(c = -1\), the function is \(f(x) = x^2 - 1\)? Wait, no, let's check with another point. Wait, when \(x = 2\), \(f(2) = 3\). Let's test \(f(2) = (2)^2 - 1 = 3\), which matches. \(x = 3\): \(3^2 - 1 = 8\), which matches. \(x = -2\): \((-2)^2 - 1 = 3\), matches. \(x = -3\): \((-3)^2 - 1 = 8\), matches. Wait, but earlier I thought of \((-1,0)\), \((1,0)\) as roots? Wait, \(f(-1) = 0\) and \(f(1) = 0\), so the roots are \(x = -1\) and \(x = 1\), so factored form is \(f(x) = a(x + 1)(x - 1)\). Then expand: \(f(x) = a(x^2 - 1)\). Use \(x = 0\), \(f(0) = a(-1) = -1 \implies a = 1\). So \(f(x) = x^2 - 1\)? Wait, but when \(x = 0\), \(f(0) = -1\), which is correct. Wait, but let's check all points:

• \(x = -3\): \((-3)^2 - 1 = 9 - 1 = 8\) ✔️
• \(x = -2\): \(4 - 1 = 3\) ✔️
• \(x = -1\): \(1 - 1 = 0\) ✔️
• \(x = 0\): \(0 - 1 = -1\) ✔️
• \(x = 1\): \(1 - 1 = 0\) ✔️
• \(x = 2\): \(4 - 1 = 3\) ✔️
• \(x = 3\): \(9 - 1 = 8\) ✔️

Wait, so the quadratic is \(f(x) = x^2 - 1\)? But wait, the table has symmetry around \(x = 0\), since \(f(-3) = f(3)\), \(f(-2) = f(2)\), \(f(-1) = f(1)\), so the vertex is at \(x = 0\), which means \(b = 0\), so the function is even, which matches \(f(x) = x^2 - 1\).

Answer:

\(f(x) = x^2 - 1\)