Practice 1 Solutions

Given the graph, determine a, h, and k. Then write the equation.

$a = 3, h = - 1, k = 2 x = 3 {(y - 2)}^{2} - 1$

Note

Remember, you also practiced reading a graph in Lesson 18.

Write a verbal description of the parabola given the equation: $y = - \frac{2}{5} {(x + 5)}^{2} + 3$

The parabola opens down and is stretched because $a = - \frac{2}{5}$ .

The axis of symmetry is $x = - 5$ . The vertex is $(- 5, 3)$ .

Write the equation of the parabola in vertex form. Name the axis of symmetry and vertex.

$x = y^{2} + 8 y - 1$

$y^{2} + 8 y - 1 = 0 y^{2} + 8 y = 1 y^{2} + 8 y + {(\frac{8}{2})}^{2} = 1 + {(\frac{8}{2})}^{2} {(y + 4)}^{2} = 1 + {(4)}^{2} {(y + 4)}^{2} = 1 + 16 {(y + 4)}^{2} = 17$

$x = {(y + 4)}^{2} - 17 AoS : y = - 4 vertex : (- 17, - 4)$

Note

Since $a = 1,$ there is no GCF to factor out before completing the square.

$3 x^{2} + y = 6 x - 2$

$y = - 3 x^{2} + 6 x - 2 - 3 x^{2} + 6 x - 2 = 0 - 3 x^{2} + 6 x = 2 - 3 (x^{2} - 2 x + {(- \frac{2}{2})}^{2}) = 2 + - 3 {(- \frac{2}{2})}^{2} - 3 {(x - 1)}^{2} = 2 - 3 {(1)}^{2} - 3 {(x - 1)}^{2} = 2 - 3 - 3 {(x - 1)}^{2} = - 1$

$y = - 3 {(x - 1)}^{2} + 1 AoS : x = 1 vertex : (1, 1)$

$y = \frac{1}{3} x^{2} + 2 x - 9$

$0 = \frac{1}{3} x^{2} + 2 x - 9 \frac{1}{3} x^{2} + 2 x = 9 \frac{1}{3} (x^{2} + 6 x + {(\frac{6}{2})}^{2}) = 9 + \frac{1}{3} {(\frac{6}{2})}^{2} \frac{1}{3} {(x + 3)}^{2} = 9 + \frac{1}{3} {(3)}^{2} \frac{1}{3} {(x + 3)}^{2} = 9 + \frac{1}{3} (9) \frac{1}{3} {(x + 3)}^{2} = 9 + 3 \frac{1}{3} {(x + 3)}^{2} = 12$

$y = \frac{1}{3} {(x + 3)}^{2} - 12 AoS : x = - 3 vertex : (- 3, - 12)$

$x = 2 y^{2} + 6 y - 7$

$\begin{array}{rcl} 2 y^{2} + 6 y - 7 & = & 0 \\ 2 y^{2} + 6 y & = & 7 \\ 2 (y^{2} + 3 y) & = & 7 \\ 2 (y^{2} + 3 y + {(\frac{3}{2})}^{2}) & = & 7 + 2 {(\frac{3}{2})}^{2} \\ 2 {(y + \frac{3}{2})}^{2} & = & 7 + 2 (\frac{9}{4}) \\ 2 {(y + \frac{3}{2})}^{2} & = & \frac{14}{2} + \frac{9}{2} \\ 2 {(y + \frac{3}{2})}^{2} & = & \frac{23}{2} \end{array}$

$x = 2 {(y + \frac{3}{2})}^{2} - \frac{23}{2} AoS : y = - \frac{3}{2} vertex : (- \frac{23}{2}, - \frac{3}{2})$

Write the equation of the equation of the parabola in vertex form. Then graph.

$2 x^{2} - y + 4 = 8 x$

$y = 2 x^{2} - 8 x + 4 2 x^{2} - 8 x = - 4 2 (x^{2} - 4 x + {(- \frac{4}{2})}^{2}) = - 4 + 2 {(- \frac{4}{2})}^{2} 2 {(x - 2)}^{2} = - 4 + 2 {(- 2)}^{2} 2 {(x - 2)}^{2} = - 4 + 2 (4) 2 {(x - 2)}^{2} = 4$

$y = 2 {(x - 2)}^{2} - 4$

$x = y^{2} + 10 y - 5$

$y^{2} + 10 y - 5 = 0 y^{2} + 10 y = 5 y^{2} + 10 y + {(\frac{10}{2})}^{2} = 5 + {(\frac{10}{2})}^{2} {(y + 5)}^{2} = 5 + {(5)}^{2} {(y + 5)}^{2} = 30$

$x = {(y + 5)}^{2} - 30$

Note

If you adjust the scale of the graph (or use technology) it is possible your graph will look more like this:

$x + y^{2} = 2 y$

$x = - y^{2} + 2 y - y^{2} + 2 y = 0 - (y^{2} - 2 y + {(\frac{- 2}{2})}^{2}) = 0 + {(\frac{- 2}{2})}^{2} - {(y - 1)}^{2} = 1 - {(y - 1)}^{2} - 1 = 0$

$x = - 2 {(y - \frac{1}{2})}^{2} + \frac{1}{2}$

$x = - {(y - 1)}^{2} - 1$

$3 x^{2} + y = 12 x$

$y = - 3 x^{2} + 12 x - 3 (x^{2} - 4 x) = 0 - 3 (x^{2} - 4 x + {(- \frac{4}{2})}^{2}) = 0 + - 3 {(- \frac{4}{2})}^{2} - 3 {(x - 2)}^{2} = - 3 {(2)}^{2} - 3 {(x - 2)}^{2} = - 3 (4) - 3 {(x - 2)}^{2} = - 12$

$y = - 3 {(x - 2)}^{2} + 12$

Write the equation of the parabola given three points.

$(4, 0), (2, 2), (- 4, - 1)$ when $x = a y^{2} + b y + c$ .

$(4, 0) a {(0)}^{2} + b (0) + c = 4 c = 4 (2, 2) a {(2)}^{2} + b (2) + c = 2 4 a + 2 b + c = 2 4 a + 2 b + 4 = 2 4 a + 2 b = - 2 (- 4, - 1) a {(- 1)}^{2} + b (- 1) + c = - 4 a - b + c = - 4 a - b + 4 = - 4 a - b = - 8$

$x = - 3 y^{2} + 5 y + 4$

Note

Remember, you can use repeated substitution for any value you have solved for in any step of the problem. Hint: Start with the point $(4, 0) .$

$(- 2, 10), (2, - 2), (3, 5)$ when $y = a x^{2} + b x + c$ .

$(- 2, 10) a {(- 2)}^{2} + b (- 2) + c = 10 4 a - 2 b + c = 10 (2, - 2) a {(2)}^{2} + b (2) + c = - 2 4 a + 2 b + c = - 2 (3, 5) a {(3)}^{2} + b (3) + c = 5 9 a + 3 b + c = 5$

$4 a - 2 b + c = 10 (4 a + 2 b + c = - 2) (- 1) \Rightarrow + - 4 a - 2 b - c = 2 - 4 b = 12 b = - 3$

$4 a - 2 b + c = 10 (9 a + 3 b + c = 5) (- 1) \Rightarrow + - 9 a - 3 b - c = - 5 - 5 a - 5 b = 5 - 5 a - 5 (- 3) = 5 - 5 a + 15 = 5 - 5 a = - 10 a = 2 4 (2) - 2 (- 3) + c = 10 8 + 6 + c = 10 c = - 4$

$y = 2 x^{2} - 3 x - 4$

Note

Alternate method using repeating substitution:

$4 a - 2 b + c = 10 + 4 a + 2 b + c = - 2 8 a + 2 c = 8 4 a + c = 4 c = 4 - 4 a$

$4 a + 2 b + (4 - 4 a) = - 2 2 b + 4 = - 2 2 b = - 6 b = - 3$

$9 a + 3 (- 3) + (4 - 4 a) = 5 9 a - 9 + 4 - 4 a = 5 5 a = 10 a = 2 c = 4 - 4 (2) c = - 4$

$(- 1, 17), (1, - 3), (3, 25)$ when $y = a x^{2} + b x + c$ .

$(- 1, 17) a {(- 1)}^{2} + b (- 1) + c = 17 a - b + c = 17 (1, - 3) a {(1)}^{2} + b (1) + c = - 3 a + b + c = - 3 (3, 25) a {(3)}^{2} + b (3) + c = 25 9 a + 3 b + c = 25$

$(a - b + c = 17) (- 1) \Rightarrow - a + b - c = - 17 + a + b + c = - 3 2 b = - 20 b = - 10$

$(a - b + c = 17) (- 1) \Rightarrow - a + b - c = - 17 + 9 a + 3 b + c = 25 8 a + 4 b = 8$

$8 a + 4 (- 10) = 8 8 a - 40 = 8 8 a = 48 a = 6 (6) + (- 10) + c = - 3 - 4 + c = - 3 c = 1$

$y = 6 x^{2} - 10 x + 1$

$(- 6, - 4), (- 10, - 2), (15, 3)$ when $x = a y^{2} + b y + c$ .

$(- 6, - 4) a {(- 4)}^{2} + b (- 4) + c = - 6 16 a - 4 b + c = - 6 (- 10, - 2) a {(- 2)}^{2} + b (- 2) + c = - 10 4 a - 2 b + c = - 10 (15, 3) a {(3)}^{2} + b (3) + c = 15 9 a + 3 b + c = 15$

$16 a - 4 b + c = - 6 (4 a - 2 b + c = - 10) (- 1) \Rightarrow + - 4 a + 2 b - c = 10 12 a - 2 b = 4 9 a + 3 b + c = 15 (4 a - 2 b + c = - 10) (- 1) \Rightarrow + - 4 a + 2 b - c = 10 5 a + 5 b = 25$

$(12 a - 2 b = 4) (5) \Rightarrow 60 a - 10 b = 20 (5 a + 5 b = 25) (2) \Rightarrow + 10 a + 10 b = 50 70 a = 70 a = 1 5 (1) + 5 b = 25 5 + 5 b = 25 5 b = 20 b = 4 9 (1) + 3 (4) + c = 15 9 + 12 + c = 15 c = - 6$

$x = y^{2} + 4 y - 6$

Describe how you know a parabola will open left/right rather than up/down.

Sample: The parabola will open left/right if the equation is in the form $x = y^{2}$ .

Natalee sets a volleyball to her hitter. The hitter wants to time her approach to hit the ball at its maximum height. The ball’s route can be modeled by the equation: $h (t) = - 16 t^{2} + 16 t + 7$ . What is the maximum height the ball will reach and how long after the ball is set will it reach that height?

$Vertex : (time, \max height) - 16 t^{2} + 16 t + 7 = 0 - 16 (t^{2} - t + {(- \frac{1}{2})}^{2}) = - 7 + - 16 {(- \frac{1}{2})}^{2} - 16 {(t - \frac{1}{2})}^{2} = - 7 - 16 (\frac{1}{4}) - 16 {(t - \frac{1}{2})}^{2} = - 11 h (t) = - 16 {(t - \frac{1}{2})}^{2} + 11 (\frac{1}{2}, 11)$

The ball will reach a maximum height of 11 feet 0.5 seconds after it is set.

The perimeter of a rectangular fence is 40 meters. What are the dimensions and the maximum area?

$P = 2 l + 2 w 40 = 2 l + 2 w 20 = l + w l = - w + 20 A = l w A = w (- w + 20) - w^{2} + 20 w = 0 - (w^{2} - 20 w + {(- \frac{20}{2})}^{2}) = 0 - {(- \frac{20}{2})}^{2} - {(w - 10)}^{2} = - 100 A = - {(w - 10)}^{2} + 100 Vertex : (10, 100) w = 10 l = - (10) + 20$

The dimensions are 10 meters by 10 meters. The maximum area is 100 square meters.

The Hawkins family purchased 180 feet of fence to enclose a rectangular area of their yard. They plan on using an existing wall for one side of the rectangular enclosure. Find the dimensions and maximum area of the enclosure.

$P = l + 2 w 180 = l + 2 w l = - 2 w + 180 A = l w A = w (- 2 w + 180) A = - 2 w^{2} + 180 w - 2 w^{2} + 180 w = 0 - 2 (w^{2} - 90 w + {(- \frac{90}{2})}^{2}) = 0 - 2 {(- \frac{90}{2})}^{2} - 2 {(w - 45)}^{2} = - 4050 A = - 2 {(w - 45)}^{2} + 4050 Vertex : (45, 4050) w = 45, l = - 2 (45) + 180$

The width is 45 feet, the length is 90 feet. The maximum area is 4050 square feet.

Note

Review the image for Example 5 if you are uncertain how to begin this problem.