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A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find the maximum height reached by the rocket, to the nearest tenth of a foot.
y, equals, minus, 16, x, squared, plus, 164, x, plus, 90
y=−16x
2
+164x+90
Answer
Attempt 1 out of 5

Answer 1

To find the maximum height reached by the rocket, we need to analyze the quadratic equation given by the formula:

\[ y = -16x^2 + 164x + 90 \]

Since this is a quadratic equation in the standard form \( ax^2 + bx + c \), we can determine the maximum height by finding the vertex of the parabola. The vertex \( x \)-coordinate for a quadratic equation is given by the formula:

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

In our equation, \( a = -16 \) and \( b = 164 \). Plugging these values into the formula:

\[ x = -\frac{164}{2(-16)} \]
\[ x = -\frac{164}{-32} \]
\[ x = \frac{164}{32} \]
\[ x = 5.125 \]

Now, we will substitute this value of \( x \) back into the original equation to find the maximum height \( y \):

\[ y = -16(5.125)^2 + 164(5.125) + 90 \]

Calculating \( (5.125)^2 \):

\[ (5.125)^2 = 26.265625 \]

Now substituting this back into the equation for \( y \):

\[
y = -16(26.265625) + 164(5.125) + 90
\]

Calculating \( -16(26.265625) \):

\[
-16(26.265625) = -420.25
\]

Calculating \( 164(5.125) \):

\[
164(5.125) = 839.5
\]

Now putting everything together:

\[
y = -420.25 + 839.5 + 90
\]
\[
y = -420.25 + 929.5
\]
\[
y = 509.25
\]

Thus, the maximum height reached by the rocket is approximately:

\[
\boxed{509.3} \text{ feet (to the nearest tenth)}
\]