Respuesta :
Using quadratic function concepts, it is found that:
1) The equation is [tex]y = -0.08x^2 + 4x[/tex]
2) The minimum distance is of 2.64 yards, and the maximum is of 47.36 yards.
A quadratic equation has the following format:
[tex]y = ax^2 + bx + c[/tex]
In this problem:
- The ball was kicked from the ground, thus [tex]y(0) = 0[/tex], which means that c = 0.
Item 1:
The vertex is: [tex]V(x_v,y_v)[/tex], in which:
[tex]x_v = -\frac{b}{2a}[/tex]
[tex]y_v = -\frac{\Delta}{4a} = -\frac{b^2 - 4ac}{4a}[/tex]
In this problem, it is (25,50). Thus:
[tex]-\frac{b}{2a} = 25[/tex]
[tex]-b = 50a[/tex]
[tex]b = -50a[/tex]
[tex]-\frac{b^2 - 4ac}{4a} = 50[/tex]
[tex]-b^2 = 200a[/tex]
[tex]-(50a)^2 = 200a[/tex]
[tex]-2500a^2 - 200a = 0[/tex]
[tex]200a(-12.5a - 1) = 0[/tex]
Then, as [tex]a \neq 0[/tex]
[tex]a = -\frac{1}{12.5} = -0.08[/tex]
[tex]b = -50a = -50(-0.08) = 4[/tex]
Thus, the equation, in standard form, is:
[tex]y = -0.08x^2 + 4x[/tex]
Item 2:
The distances are the values of x for which:
[tex]y = 10[/tex]
Then
[tex]-0.08x^2 + 4x = 10[/tex]
[tex]-0.08x^2 + 4x - 10 = 0[/tex]
Which is a quadratic equation with [tex]a = -0.08, b = 4, c = -10[/tex]. Then:
[tex]\Delta = 4^2 - 4(-0.08)(-10) = 12.8[/tex]
[tex]x_{1} = \frac{-4 + \sqrt{12.8}}{2(-0.08)} = 2.64[/tex]
[tex]x_{2} = \frac{-4 - \sqrt{12.8}}{2(-0.08)} = 47.36[/tex]
The minimum distance is of 2.64 yards, and the maximum is of 47.36 yards.
A similar problem is given at https://brainly.com/question/24713268
