Question

An object is thrown into the air. Use GeoGebra to graph the situation using the formula y = -4t^2 + 10t. What is the maximum height the object will reach in feet?

Answer 1

The given formula for the height of the object at time \( t \) is

\[
y(t) = -4t^2 + 10t.
\]

This is a quadratic equation in the standard form \( y = at^2 + bt + c \), where \( a = -4 \), \( b = 10 \), and \( c = 0 \). The graph of this equation is a parabola that opens downward (since \( a < 0 \)).

To find the maximum height of the object, we need to determine the vertex of the parabola. The time \( t \) at which the maximum height occurs can be found using the vertex formula:

\[
t = -\frac{b}{2a}.
\]

Substituting the values of \( a \) and \( b \):

\[
t = -\frac{10}{2 \cdot -4} = \frac{10}{8} = \frac{5}{4} = 1.25 \text{ seconds}.
\]

Next, we will substitute \( t = 1.25 \) back into the height equation to find the maximum height \( y \):

\[
y(1.25) = -4(1.25)^2 + 10(1.25).
\]

Calculating \( (1.25)^2 = 1.5625 \):

\[
y(1.25) = -4(1.5625) + 12.5
\]

\[
y(1.25) = -6.25 + 12.5 = 6.25 \text{ feet}.
\]

Thus, the maximum height that the object will reach is

\[
\boxed{6.25} \text{ feet}.
\]