Question

1. applied research the amounts (in billions of dollars) the u.s. federal government spent on applied research from 2013 through 2018 can be approximated by the model ( y = 0.0729t^2 - 0.526t + 24.34 ) where ( t ) represents the year, with ( t = 13 ) corresponding to 2013. (source: national center for science and engineering statistics) a. use a graphing utility to graph the model.

Explanation:

for Graphing the Model

Step 1: Identify the Function Type

The model \( y = 0.0729t^2 - 0.526t + 24.34 \) is a quadratic function (since the highest power of \( t \) is 2). The general form of a quadratic function is \( y = ax^2+bx + c \), here \( a = 0.0729 \), \( b=- 0.526 \), \( c = 24.34 \) and \( x=t \). Since \( a=0.0729>0 \), the parabola opens upwards.

Step 2: Determine the Domain

We know that \( t \) represents the year, with \( t = 13 \) corresponding to 2013 and we are considering the years from 2013 through 2018. So when \( t = 13 \) (2013), \( t=14 \) (2014), \( t = 15 \) (2015), \( t=16 \) (2016), \( t = 17 \) (2017), \( t=18 \) (2018). So the domain of \( t \) for our graph is \( t\in[13,18] \) (we can also consider a slightly larger domain for better visualization, but the key interval is from 13 to 18).

Step 3: Use a Graphing Utility

To graph the function:

• If using a graphing calculator (like TI - 84):
• Press Y= and enter the function \( Y_1=0.0729X^2 - 0.526X+24.34 \) (we use \( X \) instead of \( t \) as the variable in the calculator).
• Set the window settings. For the \( X \)-axis (representing \( t \)): set the minimum value as 12 (to see a bit before 2013), maximum as 19 (to see a bit after 2018), and an appropriate scale (e.g., scale = 1). For the \( Y \)-axis (representing \( y \), the amount in billions of dollars): we can first find the minimum and maximum values of \( y \) in the interval \( t\in[13,18] \) by plugging in the endpoints.
• When \( t = 13 \): \( y=0.0729\times13^2-0.526\times13 + 24.34=0.0729\times169-6.838 + 24.34=12.3201-6.838 + 24.34 = 29.8221\)
• When \( t = 18 \): \( y=0.0729\times18^2-0.526\times18+24.34=0.0729\times324-9.468 + 24.34=23.6196-9.468+24.34 = 38.4916\)
• So we can set the \( Y \)-axis minimum as 25, maximum as 40, and scale = 1.
• Then press GRAPH to see the graph of the quadratic function over the interval \( t\in[13,18] \).
• If using an online graphing utility (like Desmos):
• Go to the Desmos website (https://www.desmos.com/calculator).
• In the input box, type \( y = 0.0729t^2-0.526t + 24.34 \).
• To set the domain for \( t \), we can use the graph settings or just observe the graph for \( t \) values from 13 to 18. The graph will be a parabola opening upwards, and we can see how the value of \( y \) (spending on applied research) changes as \( t \) (the year) increases from 13 (2013) to 18 (2018).

Since the question only asks to use a graphing utility to graph the model, the process involves recognizing the function type, determining the appropriate domain, and then using a graphing tool (calculator or online) to plot \( y = 0.0729t^2-0.526t + 24.34 \) for \( t\in[13,18] \). The graph will be a parabola opening upwards with vertex at \( t=-\frac{b}{2a}=-\frac{- 0.526}{2\times0.0729}=\frac{0.526}{0.1458}\approx3.61 \), but since our domain is \( t\in[13,18] \), the function is increasing on this interval (because the vertex is at \( t\approx3.61 \) and the parabola opens upwards, so for \( t>3.61 \), the function is increasing).

Answer:

(for the graphing part)
To graph \( y = 0.0729t^2-0.526t + 24.34 \) (where \( t\in[13,18] \) for years 2013 - 2018):

1. Recognize it is a quadratic function (\( a = 0.0729>0 \), opens upward).
2. Use a graphing utility (calculator/Desmos):

• For a calculator (e.g., TI - 84): Enter \( Y_1=0.0729X^2 - 0.526X + 24.34 \), set window \( X:[12,19] \), \( Y:[25,40] \), then graph.
• For Desmos: Enter \( y = 0.0729t^2-0.526t + 24.34 \), observe for \( t\in[13,18] \). The graph is an upward - opening parabola, increasing on \( [13,18] \) (since vertex \( t\approx3.61<13 \)).