To answer the questions related to the football's path modeled by the equation \( h(x) = -130x(x - 60) \), we first need to analyze the equation.
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Finding the y-intercept: The y-intercept occurs when \( x = 0 \). \[ h(0) = -130(0)(0 - 60) = 0 \] Therefore, the y-intercept is \( (0, 0) \).
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Finding the time when the football is at the highest height: The given equation \( h(x) = -130x(x - 60) \) is a quadratic equation in the form \( h(x) = ax^2 + bx + c \). To find the vertex (the highest point for this downward-opening parabola), we use the formula \( x = -\frac{b}{2a} \).
First, we need to rewrite the equation in standard form: \[ h(x) = -130x^2 + 7800x \] Here, \( a = -130 \) and \( b = 7800 \).
Now, we can find \( x \): \[ x = -\frac{7800}{2(-130)} = \frac{7800}{260} = 30 \] So, the football is at its highest height after 30 seconds.
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Finding the horizontal distance the ball will travel before hitting the ground: The ball hits the ground when \( h(x) = 0 \). So, we need to solve the equation: \[ -130x(x - 60) = 0 \] This gives us two solutions: \[ -130x = 0 \quad \text{or} \quad (x - 60) = 0 \] Thus, \( x = 0 \) or \( x = 60 \).
The ball will have traveled 60 yards before hitting the ground.
Now we can summarize the answers:
- Y-intercept: \( (0, 0) \)
- Time at highest height: 30 seconds
- Distance before hitting the ground: 60 yards
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