Question

the table represents a quadratic function. write an equation of the function in standard form.

x	4	5	6	7
f(x)	3	2	3	6

the function is f(x) = □.

Explanation:

Step1: Recall quadratic - function standard form

The standard form of a quadratic function is $f(x)=ax^{2}+bx + c$.

Step2: Substitute points into the equation

Using the points $(4,3)$, $(5,2)$, and $(6,3)$:
For $(4,3)$: $3=a(4)^{2}+b(4)+c=16a + 4b + c$.
For $(5,2)$: $2=a(5)^{2}+b(5)+c=25a+5b + c$.
For $(6,3)$: $3=a(6)^{2}+b(6)+c=36a+6b + c$.

Step3: Subtract equations to eliminate $c$

Subtract the first - equation from the second:
$(25a + 5b + c)-(16a + 4b + c)=2 - 3$,
$25a+5b + c - 16a - 4b - c=-1$,
$9a + b=-1$.
Subtract the second equation from the third:
$(36a + 6b + c)-(25a+5b + c)=3 - 2$,
$36a+6b + c - 25a - 5b - c = 1$,
$11a + b=1$.

Step4: Solve the system of equations for $a$ and $b$

Subtract the equation $9a + b=-1$ from $11a + b=1$:
$(11a + b)-(9a + b)=1-(-1)$,
$11a + b - 9a - b=2$,
$2a=2$, so $a = 1$.
Substitute $a = 1$ into $9a + b=-1$:
$9(1)+b=-1$,
$b=-1 - 9=-10$.

Step5: Find $c$

Substitute $a = 1$ and $b=-10$ into $16a + 4b + c=3$:
$16(1)+4(-10)+c=3$,
$16-40 + c=3$,
$c=3 + 24=27$.

Answer:

$f(x)=x^{2}-10x + 27$