Question

An upward parabola vertex at (2, 1) and passes through (7, 6), (-4, 8), and intercepts the y-axis at 1.9 units.

a. f(x)=−0.2(x−2) ^2 +1
b. f(x)=0.2(x−2)^2+1
c. f(x)=−0.2(x−2) ^2 −1
d. f(x)=0.2(x−2)^ −1

Answer 1

The trajectory of a projectile follows a parabolic path, and its equation can be expressed as y = ax + bx², where a and b are constants. This equation can be derived by solving the equations for the x and y positions of the projectile and simplifying the result.

Step-by-step explanation:

The trajectory of a projectile follows a parabolic path. To obtain the equation of this trajectory, we can solve the equations for the x and y positions of the projectile. The x position can be expressed as x = V0t, where V0 is the initial horizontal velocity and t is time. The y position can be expressed as y = V0t - (1/2)gt², where g is the acceleration due to gravity. Substituting the value of x into the equation for y, we get y = (V0/V0²) x - (1/2)g(x/V0)². Simplifying this equation gives us the standard form of a quadratic equation, y = ax + bx², where a = (V0/V0²) and b = -(1/2)g/V0². Therefore, the trajectory equation for a projectile is of the form y = ax + bx².