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To determine the domain and range of the function \( y(x) = -4x^2 + 8x + 12 \), we need to analyze the behavior of this quadratic function.

### Step 1: Understanding the Function
The function given is:
\[
y(x) = -4x^2 + 8x + 12
\]
This is a quadratic function with a negative leading coefficient (\(-4\)), indicating it opens downwards. This means the function has a maximum point.

### Step 2: Domain
Since \( x \) represents time in seconds after the ball is launched, \( x \geq 0 \). The domain is determined by the time interval from launch until the ball hits the ground (when \( y = 0 \)).

To find the point where the ball hits the ground, set \( y(x) = 0 \) and solve for \( x \):
\[
-4x^2 + 8x + 12 = 0
\]
Dividing by -4 to simplify:
\[
x^2 - 2x - 3 = 0
\]
Factoring:
\[
(x - 3)(x + 1) = 0
\]
Thus, \( x = 3 \) or \( x = -1 \). Since time \( x \) cannot be negative, we only consider \( x = 3 \).

So, the domain is:
\[
0 \leq x \leq 3
\]

### Step 3: Range
To find the range, we need to determine the maximum height the ball reaches. This occurs at the vertex of the parabola.

The formula for the \( x \)-coordinate of the vertex of \( y = ax^2 + bx + c \) is:
\[
x = -\frac{b}{2a}
\]
For our function, \( a = -4 \) and \( b = 8 \):
\[
x = -\frac{8}{2 \times -4} = 1
\]

Now, substitute \( x = 1 \) into \( y(x) \) to find the maximum height:
\[
y(1) = -4(1)^2 + 8(1) + 12 = -4 + 8 + 12 = 16
\]

The highest point reached is 16 yards, and the lowest point is when the ball is on the ground, which is \( y = 0 \).

So, the range is:
\[
0 \leq y \leq 16
\]

### Final Answer
- **Domain:** \( 0 \leq x \leq 3 \)
- **Range:** \( 0 \leq y \leq 16 \)

Would you like further explanation on any part of this solution, or do you have questions?

Here are some additional questions to deepen your understanding:
1. What does the vertex of the quadratic function represent in this context?
2. How would the domain and range change if the initial height or launch speed were different?
3. Why does the function have a maximum point rather than a minimum?
4. How can we verify that the ball hits the ground exactly when \( x = 3 \)?
5. How would the range be affected if the ball was launched with a different speed?

**Tip:** When solving for the domain and range of a physical problem, keep in mind that context (like time or distance) can limit possible values.

A ball is launched from the top of a tower 12 yards above the ground. At the moment that the ball is launched, it is traveling upwards at a speed of 8 yards per second. The function y(x) = -4x^2 + 8x + 12 describes the height of the ball in yards after x seconds. Find the domain and range of the function. Use inequality notation.

Explore how to find the domain and range of the quadratic function y(x) = -4x^2 + 8x + 12, which models the height of a launched ball. Includes step-by-step guidance on solving for when the ball hits the ground and reaching maximum height using the vertex formula.

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