Question

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 thrown into the air, we can analyze the quadratic function given by the equation:

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

This is a parabola that opens downwards (due to the negative coefficient of \(t^2\)). The maximum height occurs at the vertex of the parabola.

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

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

where \(a\) and \(b\) are the coefficients from the standard quadratic form \(y = at^2 + bt + c\). In this case, \(a = -4\) and \(b = 10\).

Plugging in these values:

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

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

\[
y = -4(1.25)^2 + 10(1.25)
\]
\[
y = -4(1.5625) + 12.5
\]
\[
y = -6.25 + 12.5 = 6.25 \text{ feet}
\]

Therefore, the maximum height the object will reach is **6.25 feet**.