Question

1. the path of a projectile catapulted from a cliff 20 m tall is modeled by the equation y = -5x² + 40x + 20. graph the equation. find the maximum height, in meters, reached by the projectile.

160
y
120
80
40
0 0 2 4 6 8 x

Explanation:

Step1: Recall vertex formula for parabola

For a quadratic equation \( y = ax^2 + bx + c \), the x - coordinate of the vertex is given by \( x=-\frac{b}{2a} \). In the equation \( y=-5x^{2}+40x + 20 \), we have \( a=-5 \), \( b = 40 \) and \( c = 20 \).

Step2: Calculate x - coordinate of vertex

Substitute \( a=-5 \) and \( b = 40 \) into the formula \( x=-\frac{b}{2a} \).
\( x=-\frac{40}{2\times(-5)}=-\frac{40}{-10} = 4 \)

Step3: Calculate y - coordinate of vertex (maximum height)

Substitute \( x = 4 \) into the equation \( y=-5x^{2}+40x + 20 \).
\( y=-5\times(4)^{2}+40\times4 + 20=-5\times16+160 + 20=-80 + 160+20=100 \)

Answer:

The maximum height reached by the projectile is 100 meters.