Question

solving problems with quadratic functions: tutorial
photo necessities produces camera cases. they have found that the cost, c(x), of making x camera cases is a quadratic function in terms of x.
the company also discovered that it costs $23 to produce 2 camera cases, $103 to produce 4 camera cases, and $631 to produce 10 camera cases.
complete the function that represents the cost, c(x), to produce x camera cases.
c(x) =
enter the correct answer.

Explanation:

Step1: Define quadratic cost function

Let $c(x) = ax^2 + bx + c_0$, where $a$, $b$, $c_0$ are constants.

Step2: Set up system of equations

Substitute $(x, c(x))$ pairs:

1. For $x=2$, $c(2)=23$: $4a + 2b + c_0 = 23$
2. For $x=4$, $c(4)=103$: $16a + 4b + c_0 = 103$
3. For $x=10$, $c(10)=631$: $100a + 10b + c_0 = 631$

Step3: Eliminate $c_0$ (Eq2-Eq1)

Subtract Eq1 from Eq2:
$$(16a + 4b + c_0) - (4a + 2b + c_0) = 103 - 23$$
$$12a + 2b = 80 \implies 6a + b = 40 \quad \text{(Eq4)}$$

Step4: Eliminate $c_0$ (Eq3-Eq2)

Subtract Eq2 from Eq3:
$$(100a + 10b + c_0) - (16a + 4b + c_0) = 631 - 103$$
$$84a + 6b = 528 \implies 14a + b = 88 \quad \text{(Eq5)}$$

Step5: Solve for $a$ (Eq5-Eq4)

Subtract Eq4 from Eq5:
$$(14a + b) - (6a + b) = 88 - 40$$
$$8a = 48 \implies a = 6$$

Step6: Solve for $b$ (Substitute $a$ to Eq4)

Substitute $a=6$ into Eq4:
$$6(6) + b = 40$$
$$36 + b = 40 \implies b = 4$$

Step7: Solve for $c_0$ (Substitute $a,b$ to Eq1)

Substitute $a=6$, $b=4$ into Eq1:
$$4(6) + 2(4) + c_0 = 23$$
$$24 + 8 + c_0 = 23 \implies c_0 = -9$$

Step8: Form final cost function

Substitute $a=6$, $b=4$, $c_0=-9$ into $c(x)$.

Answer:

$c(x) = 6x^2 + 4x - 9$