Problem 37: The function $y = -0.024x^2 + 0.55x$ models the height $y$, in feet, of a frog's jump, where $x$ is the horizontal distance, in feet, from the start of the jump. We need to find how far the frog jumped and how high it went. Problem 38: The temperature $T$ of a metal part in degrees Fahrenheit $x$ minutes after the machine is put into operation is modeled by $T = -0.005x^2 + 0.45x + 125$. We need to determine if the temperature of the part will ever reach or exceed 132°F.

1. Problem Description

Problem 37: The function $y = -0.024x^2 + 0.55x$ models the height $y$, in feet, of a frog's jump, where $x$ is the horizontal distance, in feet, from the start of the jump. We need to find how far the frog jumped and how high it went.

Problem 38: The temperature $T$ of a metal part in degrees Fahrenheit $x$ minutes after the machine is put into operation is modeled by $T = -0.005x^2 + 0.45x + 125$. We need to determine if the temperature of the part will ever reach or exceed 132°F.

2. Solution Steps

Problem 37:

To find how far the frog jumped, we need to find the horizontal distance $x$ when the height $y$ is

0. So we solve the equation $0 = -0.024x^2 + 0.55x$.

We can factor out an $x$:

$0 = x(-0.024x + 0.55)$

So $x = 0$ or $-0.024x + 0.55 = 0$.

$0.024x = 0.55$

$x = \frac{0.55}{0.024} \approx 22.92$

The frog jumped approximately 22.92 feet.

To find the maximum height, we can find the vertex of the parabola $y = -0.024x^2 + 0.55x$.

The x-coordinate of the vertex is given by $x_v = -\frac{b}{2a}$, where $a = -0.024$ and $b = 0.55$.

$x_v = -\frac{0.55}{2(-0.024)} = \frac{0.55}{0.048} \approx 11.46$

Now we find the y-coordinate of the vertex by plugging $x_v$ into the equation:

$y_v = -0.024(11.46)^2 + 0.55(11.46) \approx -0.024(131.33) + 6.303 \approx -3.152 + 6.303 \approx 3.15$

So the maximum height is approximately 3.15 feet.

Problem 38:

We want to find the maximum temperature $T$. The vertex of the parabola $T = -0.005x^2 + 0.45x + 125$ gives the maximum temperature.

$x_v = -\frac{b}{2a} = -\frac{0.45}{2(-0.005)} = \frac{0.45}{0.01} = 45$

The maximum temperature occurs at $x = 45$ minutes. Now we find the temperature at this time:

$T = -0.005(45)^2 + 0.45(45) + 125 = -0.005(2025) + 20.25 + 125 = -10.125 + 20.25 + 125 = 135.125$

Since $135.125 > 132$, the temperature of the part will exceed 132°F.

3. Final Answer

Problem 37:

Jump distance:

$x = 22.92$

Maximum height:

$y = 3.15$

Problem 38:

Yes, the temperature will exceed 132°F.

Maximum Temperature:

$T = 135.125$