Mastery Check Solutions

1. A scale model of a self-anchored suspension bridge was created on a coordinate plane so that the equation of the main eyebar chain could be determined. Determine the parabolic equation for the eyebar chain from the three points $(0, 2.5), (–5, 7), (12, 3.94)$.

$$\begin{gathered} \left(0, 2.5\right) \\ a{x}^2+bx+c=0 \\ a{\left(0\right)}^2+b\left(0\right)+c=2.5 \\ c=2.5 \\ \left(–5, 7\right) \\ a{\left(–5\right)}^2+b\left(–5\right)+2.5=7 \\ 25a–5b=4.5 \\ \left(12, 3.94\right) \\ a{\left(12\right)}^2+b\left(12\right)+2.5=3.94 \\ 144a+12b=1.44 \end{gathered}$$

$$\begin{gathered} \left(25a–5b=4.5\right)\left(12\right)\Rightarrow 300a–60b=54 \\ \left(144a+12b=1.44\right)\left(5\right)\Rightarrow \underline{+720a+60b= 0.6} \\ 1020a=61.2 \\ a=0.06 \end{gathered}$$

$\begin{array}{rcl}5\left(0.06\right)–b& =& 0.9\\ 0.3–b& =& 0.9\\ –b& =& 0.6\\ b& =& –0.6\\ & & y=0.06x^2–0.6x+2.5\end{array}$

2. Rewrite the eyebar chain equation from part A in vertex form.

$$\begin{gathered} y=0.06x^2–0.6x+2.5 \\ 0.06x^2–0.6x+2.5=0 \\ 0.06x^2–0.6x=–2.5 \\ 0.06\left(x^2–10x+{\left(\boxed{\frac{–10}{2}}\right)}^2\right)=–2.5+0.06{\left(\boxed{\frac{–10}{2}}\right)}^2 \\ 0.06{\left(x–5\right)}^2=–2.5+0.06\left(25\right) \\ 0.06{\left(x–5\right)}^2=–1 \\ 0.06{\left(x–5\right)}^2+1=0 \\ y=0.06{\left(x–5\right)}^2+1 \end{gathered}$$

3. Graph the eyebar chain using the information from parts A and B.