Question

Algebra 2 - Standard form of a quadratic function

Answer 1

Using quadratic function concepts, it is found that:

1) The equation is $y = -0.08x^2 + 4x$

2) The minimum distance is of 2.64 yards, and the maximum is of 47.36 yards.

A quadratic equation has the following format:

$y = ax^2 + bx + c$

In this problem:

• The ball was kicked from the ground, thus $y(0) = 0$, which means that c = 0.

Item 1:

The vertex is: $V(x_v,y_v)$, in which:

$x_v = -\frac{b}{2a}$

$y_v = -\frac{\Delta}{4a} = -\frac{b^2 - 4ac}{4a}$

In this problem, it is (25,50). Thus:

$-\frac{b}{2a} = 25$

$-b = 50a$

$b = -50a$

$-\frac{b^2 - 4ac}{4a} = 50$

$-b^2 = 200a$

$-(50a)^2 = 200a$

$-2500a^2 - 200a = 0$

$200a(-12.5a - 1) = 0$

Then, as $a \neq 0$

$a = -\frac{1}{12.5} = -0.08$

$b = -50a = -50(-0.08) = 4$

Thus, the equation, in standard form, is:

$y = -0.08x^2 + 4x$

Item 2:

The distances are the values of x for which:

$y = 10$

Then

$-0.08x^2 + 4x = 10$

$-0.08x^2 + 4x - 10 = 0$

Which is a quadratic equation with $a = -0.08, b = 4, c = -10$. Then:

$\Delta = 4^2 - 4(-0.08)(-10) = 12.8$

$x_{1} = \frac{-4 + \sqrt{12.8}}{2(-0.08)} = 2.64$

$x_{2} = \frac{-4 - \sqrt{12.8}}{2(-0.08)} = 47.36$  

The minimum distance is of 2.64 yards, and the maximum is of 47.36 yards.

A similar problem is given at brainly.com/question/24713268