Modeling with Linear and Quadratic Functions

1. (a) $C(h) = 40h + 50$

(b) The $40$ represents the rate ($40 per hour) — it is the slope, showing how cost increases per hour. The $50$ represents the flat fee — it is the $y$-intercept (the cost even at $0$ hours).

(c) Set $C(h) = 210$:

$40h + 50 = 210 \implies 40h = 160 \implies h = 4 \text{ hours}$

2. (a) $h(0) = -16(0)^2 + 100 = 100$ feet. This is the initial height — the ball starts at $100$ feet, the top of the building.

(b)

• $h(1) = -16(1)^2 + 100 = -16 + 100 = 84$ feet
• $h(2) = -16(2)^2 + 100 = -64 + 100 = 36$ feet

(c) Set $h(t) = 0$:

$-16t^2 + 100 = 0 \implies 16t^2 = 100 \implies t^2 = \frac{100}{16} = 6.25 \implies t = \sqrt{6.25} = 2.5 \text{ seconds}$

The ball hits the ground at $t = 2.5$ seconds.

3. (a)

• Alex: $A(w) = 15w + 20$
• Jordan: $J(w) = 20w + 5$

(b) Set equal:

$15w + 20 = 20w + 5 \implies 15 = 5w \implies w = 3 \text{ weeks}$

After $3$ weeks, both have A(3) = 15(3) + 20 = \65$.

(c) This is a linear model. Both savings functions are first-degree (each week savings increases by a fixed amount, not a fixed factor). Graphs of both would be straight lines.

4. (a) The two sides perpendicular to the barn use $2x$ feet of fencing. The remaining fencing for the side parallel to the barn is:

$80 - 2x \text{ feet}$

(b) Area $=$ width $\times$ length:

$A(x) = x(80 - 2x) = 80x - 2x^2$

(c) The function is $A(x) = -2x^2 + 80x$, so $a = -2$, $b = 80$.

$x = -\frac{b}{2a} = -\frac{80}{2(-2)} = -\frac{80}{-4} = 20 \text{ feet}$

Maximum area:

$A(20) = -2(20)^2 + 80(20) = -2(400) + 1600 = -800 + 1600 = 800 \text{ square feet}$

The maximum area is 800 sq ft when each perpendicular side is $20$ ft (and the parallel side is $80 - 2(20) = 40$ ft).

5. Check first differences ($\Delta y$) and second differences ($\Delta^2 y$):

| $x$ | $y$ | $\Delta y$ | $\Delta^2 y$ | |-----|-----|-----------|-------------| | 0 | 1 | — | — | | 1 | 4 | 3 | — | | 2 | 9 | 5 | 2 | | 3 | 16 | 7 | 2 | | 4 | 25 | 9 | 2 |

First differences are not constant (the data is not linear). Second differences are constant ($= 2$), which is the signature of a quadratic function.

The data fits $y = (x+1)^2 = x^2 + 2x + 1$. A quadratic model is appropriate.

6. $P(x) = -2x^2 + 24x - 54$, so $a = -2$, $b = 24$, $c = -54$.

(a) Price that maximizes profit:

$x = -\frac{b}{2a} = -\frac{24}{2(-2)} = -\frac{24}{-4} = \$6$

(b) Maximum profit:

$P(6) = -2(6)^2 + 24(6) - 54 = -2(36) + 144 - 54 = -72 + 144 - 54 = 18$

Maximum monthly profit is $18,000.

(c) Find roots of $-2x^2 + 24x - 54 = 0$. Divide by $-2$:

$x^2 - 12x + 27 = 0$

Factor: $(x-3)(x-9) = 0 \implies x = 3$ or $x = 9$

The company makes a positive profit when $P(x) > 0$, which occurs between the roots:

$3 < x < 9$

The company must charge between $3 and $9 per unit (exclusive) to earn a positive profit.