Question

(u5) two jedi are having a training battle. one jedi (the master) runs up a wall and does a flip in the air. his jump can be modeled by the equation: $f(x) = -0.8x^2 + 4x + 6$, where $x$ equals the time in the air and $f(x)$ is the height of his jump. the other jedi (his apprentice) swings her lightsaber at her master. the swing of the other jedis lightsaber can be measured by $g(x) = .85x + 1.75$ where $x$ is the time in seconds and $g(x)$ is the height of her lightsaber in feet. when will the lightsaber hit the jedi?

Explanation:

Step1: Set the two functions equal

To find when the lightsaber hits the Jedi, we set \( f(x) = g(x) \), so:
\( -0.8x^2 + 4x + 6 = 0.85x + 1.75 \)

Step2: Rearrange into standard quadratic form

Subtract \( 0.85x \) and \( 1.75 \) from both sides:
\( -0.8x^2 + 4x - 0.85x + 6 - 1.75 = 0 \)
Simplify:
\( -0.8x^2 + 3.15x + 4.25 = 0 \)
Multiply both sides by -100 to eliminate decimals:
\( 80x^2 - 315x - 425 = 0 \)
Divide by 5:
\( 16x^2 - 63x - 85 = 0 \)

Step3: Solve the quadratic equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 16 \), \( b = -63 \), \( c = -85 \).
First, calculate the discriminant \( D = b^2 - 4ac \):
\( D = (-63)^2 - 4(16)(-85) = 3969 + 5440 = 9409 \)
\( \sqrt{D} = 97 \)
Then,
\( x = \frac{63 \pm 97}{32} \)
We have two solutions:
\( x_1 = \frac{63 + 97}{32} = \frac{160}{32} = 5 \)
\( x_2 = \frac{63 - 97}{32} = \frac{-34}{32} = -1.0625 \) (discard since time can't be negative)

Answer:

5