Question

activity
the marine biology students use a special device to launch sensors into the ocean off piers or boats at various locations. the sensors used in shallow water and deep water are handled by different devices. after launching, their height above the water level can be modeled by these quadratic functions, where h(t) is the height, in meters, after t seconds:

• shallow-water sensor: h(t) = -4.9t² + 19.6t + 10.4
• deep-water sensor: h(t) = -4.9t² + 29.4t + 5.9

question 1
question
this equation models the height over time of a shallow-water sensor.
h(t) = -4.9t² + 19.6t + 10.4
graph the function. then mark the vertex and any intercepts that are reasonable in terms of the applicable domain and range for this situation.

Explanation:

Step1: Identify the type of quadratic function

The function \( h(t) = -4.9t^2 + 19.6t + 10.4 \) is a quadratic function in the form \( ax^2+bx + c \) with \( a=-4.9 \), \( b = 19.6 \), \( c=10.4 \). Since \( a<0 \), the parabola opens downward.

Step2: Find the vertex

The \( t \)-coordinate of the vertex of a quadratic \( ax^2+bx + c \) is given by \( t=-\frac{b}{2a} \).
Substitute \( a=-4.9 \) and \( b = 19.6 \):
\( t=-\frac{19.6}{2\times(-4.9)}=-\frac{19.6}{-9.8} = 2 \)
To find the \( h(t) \)-coordinate, substitute \( t = 2 \) into the function:
\( h(2)=-4.9\times(2)^2+19.6\times(2)+10.4=-4.9\times4 + 39.2+10.4=-19.6+39.2 + 10.4=30 \)
So the vertex is at \( (2, 30) \).

Step3: Find the y - intercept (t = 0)

Substitute \( t = 0 \) into \( h(t) \):
\( h(0)=-4.9\times0^2+19.6\times0 + 10.4=10.4 \)
So the y - intercept is \( (0, 10.4) \).

Step4: Find the t - intercept (h(t)=0)

Set \( -4.9t^2+19.6t + 10.4 = 0 \)
Multiply both sides by - 1 to get \( 4.9t^2-19.6t - 10.4 = 0 \)
Using the quadratic formula \( t=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \), where \( a = 4.9 \), \( b=-19.6 \), \( c=-10.4 \)
First, calculate the discriminant \( D=b^2-4ac=(-19.6)^2-4\times4.9\times(-10.4)=384.16 + 203.84=588 \)
\( \sqrt{D}=\sqrt{588}\approx24.25 \)
\( t=\frac{19.6\pm24.25}{2\times4.9}=\frac{19.6\pm24.25}{9.8} \)
We have two solutions:
\( t_1=\frac{19.6 + 24.25}{9.8}=\frac{43.85}{9.8}\approx4.47 \)
\( t_2=\frac{19.6-24.25}{9.8}=\frac{-4.65}{9.8}\approx - 0.47 \)
Since time \( t\geq0 \) (in the context of the problem, we can't have negative time), we consider only \( t\approx4.47 \). So the t - intercept is approximately \( (4.47, 0) \).

Step5: Determine the domain and range

The domain of \( t \) is \( t\geq0 \) (time starts at 0 when the sensor is launched). The range of \( h(t) \) is \( 0\leq h(t)\leq30 \) (since the parabola opens downward, the maximum height is at the vertex \( h = 30 \) and the minimum height is 0 when the sensor hits the water).

To graph the function:

• Plot the vertex \( (2, 30) \), the y - intercept \( (0, 10.4) \), and the t - intercept \( (4.47, 0) \).
• Draw a parabola opening downward passing through these points.

Answer:

The vertex of the function \( h(t)=-4.9t^2 + 19.6t + 10.4 \) is \( (2, 30) \), the y - intercept is \( (0, 10.4) \), and the relevant t - intercept (for \( t\geq0 \)) is approximately \( (4.47, 0) \). When graphing, plot these points and draw a downward - opening parabola through them.