Answer:
The bird will be at a ground distance of 10.04 units away.
Step-by-step explanation:
Vertex of a quadratic function:
Suppose we have a quadratic function in the following format:
[tex]f(x) = ax^{2} + bx + c[/tex]
It's vertex is the point [tex](x_{v}, y_{v})[/tex]
In which
[tex]x_{v} = -\frac{b}{2a}[/tex]
[tex]y_{v} = -\frac{\Delta}{4a}[/tex]
Where
[tex]\Delta = b^2-4ac[/tex]
If a<0, the vertex is a maximum point, that is, the maximum value happens at [tex]x_{v}[/tex], and it's value is [tex]y_{v}[/tex].
Equation for the height:
The height of the bird after x seconds is given by:
[tex]h(x) = -0.114x^2 + 2.29x + 3.5[/tex]
Which is a quadratic equation with [tex]a = -0.114, b = 2.29, c = 3.5[/tex].
When the bird is at its highest?
Quadratic equation with [tex]a < 0[/tex], and thus, at the vertex. The ground distance is the x-value of the vertex. Thus
[tex]x_{v} = -\frac{b}{2a} = -\frac{2.29}{2(-0.114)} = 10.04[/tex]
The bird will be at a ground distance of 10.04 units away.
