Mastery Check Solutions

1. Determine$f\left(\frac{1}{2}\right) \mathrm{if} f\left(–4\right)=40 \mathrm{and} f\left(x\right)=a{x}^2–5x–12.$

$\begin{array}{rcl} f\left(–4\right)& =& a{\left(–4\right)}^2–5\left(–4\right)–12 \\ 40& =& 16a+20–12\\ 40& =& 16a+8\\ 32& =& 16a\\ a& =& 2\\ & & f\left(x\right)=2x^2–5x–12\end{array}$

$$\begin{gathered} f\left(\frac{1}{2}\right)=2{\left(\frac{1}{2}\right)}^2–5\left(\frac{1}{2}\right)–12 \\ f\left(\frac{1}{2}\right)=\frac{1}{2}–\frac{5}{2}–12 \\ f\left(\frac{1}{2}\right)=–14 \end{gathered}$$

2. Fill in f(x) based on your work in part A. Then use the functions f(x), g(x), and h(x) to complete parts C, D, and E.

$f\left(x\right)={\underline{ 2x^2–5x–12 }}_{ }g\left(x\right)=\frac{4}{2x+3} h\left(x\right)=3x–12$

3. Find the sum of two functions that result in a quadratic trinomial. Show your work.

$$\begin{gathered} \left(f+h\right)\left(x\right)=f\left(x\right)+h\left(x\right) \\ \left(f+h\right)\left(x\right)=\left(2x^2–5x–12\right)+\left(3x–12\right) \\ \left(f+h\right)\left(x\right)=2x^2–2x–24 \end{gathered}$$

4. Find the product of two functions that result in a linear binomial. Show your work.

$$\begin{gathered} f\left(x\right)·g\left(x\right) \\ \left(2x^2–5x–12\right)\left(\frac{4}{2x+3}\right) \\ \left(\cancel{2x+3}\right)\left(x–4\right)\left(\frac{4}{\cancel{2x+3}}\right) \\ 4\left(x–4\right) \\ 4x–16 \end{gathered}$$

5. Evaluate the new functions in parts C and D when $x=10.$

$$\begin{gathered} \left(f+h\right)\left(10\right)=2{\left(10\right)}^2–2\left(10\right)–24 \\ \left(f+h\right)\left(10\right)=200–20–24 \\ \left(f+h\right)\left(10\right)=156 \\ \left(fg\right)\left(10\right)=4\left(10\right)–16 \\ \left(fg\right)\left(10\right)=40–16 \\ \left(fg\right)\left(10\right)=24 \end{gathered}$$