Question

An object is thrown into the air. Use GeoGebra to graph the situation using the formula y=−4t2+10t . What is the maximum height the object will reach in feet?(1 point) () feet

Answer 1

To find the maximum height of the object described by the equation \( y = -4t^2 + 10t \), we can use the vertex formula for a quadratic equation of the form \( y = ax^2 + bx + c \).

For the equation \( y = -4t^2 + 10t \):
- \( a = -4 \)
- \( b = 10 \)

The time \( t \) at which the maximum height occurs can be found using the formula:

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

Plugging in the values of \( a \) and \( b \):

\[
t = -\frac{10}{2(-4)} = \frac{10}{8} = 1.25 \text{ seconds}
\]

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

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

Calculating \( (1.25)^2 \):

\[
(1.25)^2 = 1.5625
\]

Now substituting:

\[
y = -4(1.5625) + 10(1.25)
\]

Calculating further:

\[
y = -6.25 + 12.5 = 6.25 \text{ feet}
\]

Thus, the maximum height of the object is **6.25 feet**.