Question

(u5) given the table represents a quadratic, write the equation

x	-3	-2	-1	0	1	2	3
y	21	12	5	0	-3	-4	-3

equation: y= dropdownx^2 + dropdownx + dropdown

Explanation:

Step1: Recall quadratic form

A quadratic equation is in the form \( y = ax^2 + bx + c \). We can use the table values to find \( a \), \( b \), and \( c \). Let's start with \( x = 0 \), since when \( x = 0 \), \( y = c \). From the table, when \( x = 0 \), \( y = 0 \), so \( c = 0 \). Now the equation simplifies to \( y = ax^2 + bx \).

Step2: Use another point

Let's use \( x = 1 \), \( y = -3 \). Substitute into \( y = ax^2 + bx \): \( -3 = a(1)^2 + b(1) \), so \( a + b = -3 \) (Equation 1).
Use \( x = 2 \), \( y = -4 \). Substitute: \( -4 = a(2)^2 + b(2) \), so \( 4a + 2b = -4 \) (Equation 2).

Step3: Solve the system of equations

We have \( a + b = -3 \) and \( 4a + 2b = -4 \). Let's solve Equation 1 for \( b \): \( b = -3 - a \). Substitute into Equation 2:
\( 4a + 2(-3 - a) = -4 \)
\( 4a - 6 - 2a = -4 \)
\( 2a - 6 = -4 \)
\( 2a = 2 \)
\( a = 1 \)
Now substitute \( a = 1 \) into \( b = -3 - a \): \( b = -3 - 1 = -4 \).

Step4: Verify with another point

Let's check with \( x = -1 \), \( y = 5 \). Plug into \( y = x^2 - 4x + 0 \): \( y = (-1)^2 - 4(-1) = 1 + 4 = 5 \), which matches. Check \( x = 3 \), \( y = -3 \): \( y = 3^2 - 4(3) = 9 - 12 = -3 \), which also matches.

Answer:

\( y = 1x^2 + (-4)x + 0 \) (So the coefficients are 1, -4, 0)