Practice 2 Solutions

2. Write a verbal description of the parabola given the equation: $x=\frac{1}{2}{\left(y–7\right)}^2–2$

Sample: The parabola opens right and is wide because $a=\frac{1}{2}$. The axis of symmetry is$y=7$. The vertex is $(–2, 7)$.

3. $y–4x^2=4x+3$

$$\begin{gathered} y=4x^2+4x+3 \\ 4x^2+4x=–3 \\ 4\left(x^2+x+\boxed{{\left(\frac{1}{2}\right)}^2}\right)=–3+4\boxed{{\left(\frac{1}{2}\right)}^2} \\ 4{\left(x+\frac{1}{2}\right)}^2=–3+4\left(\frac{1}{4}\right) \\ 4{\left(x+\frac{1}{2}\right)}^2=–2 \end{gathered}$$

$y=4{\left(x+\frac{1}{2}\right)}^2+2$

4. $x=–2y^2+4y–1$

$$\begin{gathered} –2y^2+4y–1=0 \\ –2\left(y^2–2y+\boxed{{\left(–\frac{2}{2}\right)}^2}\right)=1+–2\boxed{{\left(–\frac{2}{2}\right)}^2} \\ –2{\left(y–1\right)}^2=1–2\left(1\right) \\ –2{\left(y–1\right)}^2=–1 \end{gathered}$$

$x=–2{\left(y–1\right)}^2+1$

6. $y^2–\frac{1}{2}x+2=6y$

$$\begin{gathered} \frac{1}{2}x=y^2–6y+2 \\ x=2y^2–12y+4 \\ 2y^2–12y+4=0 \\ 2\left(y^2–6y+\boxed{{\left(–\frac{6}{2}\right)}^2}\right)=–4+2\boxed{{\left(–\frac{6}{2}\right)}^2} \\ 2{\left(y–3\right)}^2=–4+2\left(9\right) \\ 2{\left(y–3\right)}^2=14 \end{gathered}$$

$x=2{\left(y–3\right)}^2–14$

7. $y=\frac{ 1}{2}{x}^2+x–1$

$$\begin{gathered} \frac{1}{2}{x}^2+x–1=0 \\ \frac{1}{2}\left(x^2+2x+\boxed{{\left(\frac{2}{2}\right)}^2}\right)=1+\frac{1}{2}\boxed{{\left(\frac{2}{2}\right)}^2} \\ \frac{1}{2}{\left(x+1\right)}^2=1+\frac{1}{2} \\ \frac{1}{2}{\left(x+1\right)}^2=\frac{3}{2} \end{gathered}$$

$y=\frac{1}{2}{\left(x+1\right)}^2–\frac{3}{2}$

8. $3y^2–x+6y=2$

$$\begin{gathered} 3y^2+6y–2=x \\ 3y^2+6y–2=0 \\ 3\left(y^2+2y+\boxed{{\left(\frac{2}{2}\right)}^2}\right)=2+3\boxed{{\left(\frac{2}{2}\right)}^2} \\ 3{\left(y+1\right)}^2=2+3\left(1\right) \\ 3{\left(y+1\right)}^2=5 \end{gathered}$$

$x=3{\left(y+1\right)}^2–5$

9. $x=–\frac{1}{3}{y}^2–2y$

$$\begin{gathered} –\frac{1}{3}{y}^2–2y=0 \\ –\frac{1}{3}\left(y^2+6y+\boxed{{\left(\frac{6}{2}\right)}^2}\right)=0–\frac{1}{3}\boxed{{\left(\frac{6}{2}\right)}^2} \\ –\frac{1}{3}{\left(y+3\right)}^2=–\frac{1}{3}\left(9\right) \\ –\frac{1}{3}{\left(y+3\right)}^2=–3 \end{gathered}$$

$x=–\frac{1}{3}{\left(y+3\right)}^2+3$

10. $5x^2+y=–5x$

$$\begin{gathered} –5x^2–5x=y \\ –5x^2–5x=0 \\ –5\left(x^2+x+\boxed{{\left(–\frac{1}{2}\right)}^2}\right)=0+–5\boxed{{\left(–\frac{1}{2}\right)}^2} \\ –5{\left(x+\frac{1}{2}\right)}^2=–5\left(\frac{1}{4}\right) \\ –5{\left(x+\frac{1}{2}\right)}^2=–\frac{5}{4} \end{gathered}$$

$y=–5{\left(x+\frac{1}{2}\right)}^2+\frac{5}{4}$

11. $(1, 9.5), (0, 8), (4, 20)$ when $y=a{x}^2+bx+c$.

$$\begin{gathered} \left(0, 8\right) \\ a{\left(0\right)}^2+b\left(0\right)+c=8 \\ c=8 \\ \left(1, 9.5\right) \\ a{\left(1\right)}^2+b\left(1\right)+c=9.5 \\ a+b+c=9.5 \\ a+b+\left(8\right)=9.5 \\ a+b=1.5 \\ \left(4, 20\right) \\ a{\left(4\right)}^2+b\left(4\right)+c=20 \\ 16a+4b+c=20 \\ 16a+4b+\left(8\right)=20 \\ 16a+4b=12 \\ 4a+b=3 \end{gathered}$$

$y=0.5x^2+x+8$

12. $(–2, 0), (5, –2), (–7, 10)$ when $x=a{y}^2+by+c$.

$$\begin{gathered} \left(–2, 0\right) \\ a{\left(0\right)}^2+b\left(0\right)+c=–2 \\ c=–2 \\ \left(5, –2\right) \\ a{\left(–2\right)}^2+b\left(–2\right)+c=5 \\ 4a–2b+c=5 \\ 4a–2b+\left(–2\right)=5 \\ 4a–2b=7 \\ \left(–7, 10\right) \\ a{\left(10\right)}^2+b\left(10\right)+c=–7 \\ 100a+10b+c=–7 \\ 100a+10b+\left(–2\right)=–7 \\ 100a+10b=–5 \end{gathered}$$

$$\begin{gathered} 100a+10b=–5 \\ \left(4a–2b=7\right)\left(5\right)\Rightarrow \underline{+20a–10b=35} \\ 120a=30 \\ a=0.25 \end{gathered}$$

$$\begin{gathered} 100\left(0.25\right)+10b=–5 \\ 25+10b=–5 \\ 10b=–30 \\ b=–3 \end{gathered}$$

$x=0.25y^2–3y–2$

13. $(3, –30), (1, 4), (–1, –2)$ when$y=a{x}^2+bx+c$.

$$\begin{gathered} \left(3, –30\right) \\ a{\left(3\right)}^2+b\left(3\right)+c=–30 \\ 9a+3b+c=–30 \\ \left(1, 4\right) \\ a{\left(1\right)}^2+b\left(1\right)+c=4 \\ a+b+c+=4 \\ \left(–1, –2\right) \\ a{\left(–1\right)}^2+b\left(–1\right)+c=–2 \\ a–b+c=–2 \end{gathered}$$

$$\begin{gathered} a+b+c = 4 \\ \underline{+a–b+c=–2} \\ 2a+2c=2 \\ 9a+3b+ c=–30 \\ \left(a–b+c=–2\right)\left(3\right)\Rightarrow \underline{+ 3a–3b+3c= –6} \\ 12a+4c=–36 \end{gathered}$$

$$\begin{gathered} \left(2a+2c=2\right)\left(–2\right)\Rightarrow –4a–4c= –4 \\ \underline{+12a+4c=–36} \\ 8a=–40 \\ a=–5 \end{gathered}$$

$$\begin{gathered} 2\left(–5\right)+2c=2 \\ –10+2c=2 \\ 2c=12 \\ c=6 \\ \left(–5\right)+b+\left(6\right)=4 \\ 1+b=4 \\ b=3 \end{gathered}$$

$y=–5x^2+3x+6$

14. $(–18, –3), (–1, –1), (–15, 3)$ when $x=a{y}^2+by+c$.

$$\begin{gathered} \left(–18, –3\right) \\ a{\left(–3\right)}^2+b\left(–3\right)+c=–18 \\ 9a–3b+c=–18 \\ \left(–1, –1\right) \\ a{\left(–1\right)}^2+b\left(–1\right)+c=–1 \\ a–b+c=–1 \\ \left(–15, 3\right) \\ a{\left(3\right)}^2+b\left(3\right)+c=–15 \\ 9a+3b+c=–15 \end{gathered}$$

$$\begin{gathered} 9a–3b+c=–18 \\ \left(9a+3b+c=–15\right)\left(–1\right)\Rightarrow \underline{+ –9a–3b–c= 15} \\ –6b=–3 \\ b=0.5 \end{gathered}$$

$$\begin{gathered} 9a–3b+c=–18 \\ \left(a–b+c=–1\right)\left(–1\right)\Rightarrow \underline{+ –a+b–c= 1} \\ 8a–2b=–17 \\ 8a–2\left(0.5\right)=–17 \\ 8a–1=–17 \\ 8a=–16 \\ a=–2 \end{gathered}$$

$$\begin{gathered} \left(–2\right)–\left(0.5\right)+c=–1 \\ –2.5+c=–1 \\ c=1.5 \end{gathered}$$

$x=–2y^2+0.5y+1.5$

15. Jacob was holding a ball 2.5 feet above the ground and tossing it into the air. The height of the ball relative to the ground as a function of time could be represented by the equation: $h\left(t\right)=–16t^2+32t+2.5$ where t is time in seconds from when the ball leaves his hands and h is the height in feet the ball is above the ground. Find the maximum height of the ball and the time it occurred.

$$\begin{gathered} \mathrm{vertex}: \left(\mathrm{time}, \max \mathrm{height}\right) \\ –16t^2+32t+2.5=0 \\ –16\left(t^2–2t+\boxed{{\left(–\frac{2}{2}\right)}^2}\right)=–2.5+–16\boxed{{\left(–\frac{2}{2}\right)}^2} \\ –16{\left(t–1\right)}^2=–2.5–16\left(1\right) \\ –16{\left(t–1\right)}^2=–18.5 \\ h\left(t\right)=–16{\left(t–1\right)}^2+18.5 \\ \left(1, 18.5\right) \end{gathered}$$

After 1 second the ball reached a maximum height of 18.5 feet.  

16. The Donovans were building a rectangular chicken yard. They have 150 feet of fence. Find the dimensions of the yard and the maximum area.

$$\begin{gathered} P=2l+2w \\ 150=2l+2w \\ 75=l+w \\ l=–w+75 \\ A=lw \\ A=w\left(–w+75\right) \\ A=–w^2+75w \\ –w^2+75w=0 \\ –\left(w^2–75w+\boxed{{\left(–\frac{75}{2}\right)}^2}\right)=0–\boxed{{\left(–\frac{75}{2}\right)}^2} \\ –{\left(w–37.5\right)}^2=–1406.25 \\ A=–{\left(w–37.5\right)}^2+1406.25 \\ Vertex: \left(37.5, 1406.25\right) \\ w=37.5, l=–\left(37.5\right)+75 \end{gathered}$$

The width and length are 37.5 feet. The maximum area is 1406.25 square feet.