Question

14 multiple choice 1 point find the quadratic equation for the relationship of the horizontal distance and the height of the ball. distance (feet), x height (feet), f(x) 0 4 2 8.4 6 12.1 9 14.2 12 13.2 13 10.5 15 9.8 desmos graph interface options: y = -0.12x² + 2.11x + 4.22; y = 0.3x² - 0.4x + 07; y = 0.6x² - 0.2x + 0.7; y = -0.12x² - 0.4x + 0.4

Explanation:

Step1: Test x=0

For a quadratic equation \( y = ax^2 + bx + c \), when \( x = 0 \), \( y = c \). From the table, when \( x = 0 \), \( y = 4 \). Let's check each option:

• Option 1: \( y=- 0.12x^{2}+2.11x + 4.22 \), when \( x = 0 \), \( y = 4.22\approx4 \) (close).
• Option 2: \( y = 0.3x^{2}-0.4x + 07 \) (assuming it's \( y=0.3x^{2}-0.4x + 7 \)), when \( x = 0 \), \( y = 7

eq4 \). Eliminate.

• Option 3: \( y=0.6x^{2}-0.2x + 0.7 \), when \( x = 0 \), \( y = 0.7

eq4 \). Eliminate.

• Option 4: \( y=-0.12x^{2}-0.4x + 0.4 \), when \( x = 0 \), \( y = 0.4

eq4 \). Eliminate.

Step2: Test x=2

From the table, when \( x = 2 \), \( y = 8.4 \). For Option 1: \( y=-0.12\times(2)^{2}+2.11\times2 + 4.22=- 0.12\times4+4.22 + 4.22=-0.48 + 8.44 = 7.96\approx8.4 \) (close).

Answer:

\( y=- 0.12x^{2}+2.11x + 4.22 \) (the first option, note: there is a typo in the option, it should be \( y=-0.12x^{2}+2.11x + 4.22 \) instead of \( y - 0.12x^{2}+2.11x+4.22 \))