Targeted Review Solutions

Write the equation in standard form.

Write the equation $x = 2 y^{2} - 2 y + 2$ in the form $x = a {(y - k)}^{2} + h$ .

$2 y^{2} - 2 y + 2 = 0 2 y^{2} - 2 y = - 2 2 (y^{2} - y + {(\frac{1}{2})}^{2}) = - 2 + 2 {(\frac{1}{2})}^{2} 2 {(y - \frac{1}{2})}^{2} = - 2 + 2 (\frac{1}{4}) 2 {(y - \frac{1}{2})}^{2} = - \frac{3}{2} 2 {(y - \frac{1}{2})}^{2} + \frac{3}{2} = 0$

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

Graph problem 1.

Vertex $(1.5, 0.5)$

$a = 2$

Note

It may be more efficient to count by halves so that the parabola can be plotted more accurately.

Write the equation $x^{2} + y^{2} + 4 x + y - 4 \frac{3}{4} = 0$ in the form ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ .

$x^{2} + 4 x + y^{2} + y = 4 \frac{3}{4} (x^{2} + 4 x + {(\frac{4}{2})}^{2}) + (y^{2} + y + {(\frac{1}{2})}^{2}) = \frac{19}{4} + {(\frac{4}{2})}^{2} + {(\frac{1}{2})}^{2} {(x + 2)}^{2} + {(y + \frac{1}{2})}^{2} = \frac{19}{4} + \frac{16}{4} + \frac{1}{4} {(x + 2)}^{2} + {(y + \frac{1}{2})}^{2} = \frac{36}{4}$

${(x + 2)}^{2} + {(y + \frac{1}{2})}^{2} = 9$

Graph problem 3.

Center: $(- 2, - 0.5)$
Radius: 3

Find the distance between the vertices in Quadrant 2 and Quadrant 4 from the graph in problem 3.

Quadrant 2: (–2, 2.5)
Quadrant 4: (1, –0.5)

$d = \sqrt{{(- 2 - (1))}^{2} + {(2.5 - (- 0.5))}^{2}} d = \sqrt{{(3)}^{2} + {(3)}^{2}} d = \sqrt{18}$

$d = 3 \sqrt{2}$

Complete the square over the set of complete numbers.
$x^{2} = 22 - 6 x$

$\begin{array}{rcl} x^{2} + 6 x - 22 & = & 0 \\ x^{2} + 6 x + {(\frac{6}{2})}^{2} & = & 22 + {(\frac{6}{2})}^{2} \\ {(x + 3)}^{2} & = & 22 + 9 \\ {(x + 3)}^{2} & = & 31 \\ \sqrt{{(x + 3)}^{2}} & = & \pm \sqrt{31} \\ x + 3 & = & \pm \sqrt{31} \end{array}$

$x = - 3 \pm \sqrt{31}$

Solve $y = 2 x^{2} - 2 x + 2$ using the quadratic formula.

$a = 2, b = - 2, c = 2 x = \frac{- (- 2) \pm \sqrt{{(- 2)}^{2} - 4 (2) (2)}}{2 (2)} x = \frac{2 \pm \sqrt{4 - 16}}{4} = \frac{2 \pm \sqrt{- 12}}{4} x = \frac{2 \pm i \sqrt{12}}{4} = \frac{2 \pm 2 i \sqrt{3}}{4}$

$x = \frac{1 \pm i \sqrt{3}}{2}$

Solve $3 x^{3} - 15 x - 20 = 4 x^{2}$ by factoring under the set of complex numbers.

$\begin{array}{rcl} 3 x^{3} - 4 x^{2} - 15 x + 20 & = & 0 \\ (3 x^{3} - 4 x^{2}) + (- 15 x + 20) & = & 0 \\ x^{2} (3 x - 4) - 5 (3 x - 4) & = & 0 \\ (3 x - 4) (x^{2} - 5) & = & 0 \\ (3 x - 4) (x - \sqrt{5}) (x + \sqrt{5}) & = & 0 \end{array}$

$x = \frac{4}{3}, x = \pm \sqrt{5}$

Multiple Choice 
Use the graph to answer problems 9–10.

Determine the equation of the circle with a center in the fourth quadrant.

${(x - 8)}^{2} + {(y + 12)}^{2} = 20^{2}$
${(x + 8)}^{2} + {(y - 12)}^{2} = 20^{2}$
${(x - 8)}^{2} + {(y + 12)}^{2} = 10^{2}$
${(x + 8)}^{2} + {(y - 12)}^{2} = 10^{2}$

Q4: (+x, –y)
(8, –12)
r = 20

Note

The values of h and k represent a circle in Quadrant 2.

C, D) The value of the radius is incorrect.

Write the equation of the parabola reflected over the x-axis compared to the parent graph
$y = x^{2} .$

$x = - \frac{1}{8} {(y - 12)}^{2} + 8$
$y = - \frac{1}{4} {(x + 12)}^{2} - 8$
$x = \frac{1}{8} {(y - 12)}^{2} + 8$
$y = \frac{1}{4} {(x + 12)}^{2} - 8$

Reflected over x-axis: –a
Vertex:

$(- 12, - 8)$ $a = - \frac{1}{4}$

Note

A, C, D) These equations are not reflected over the x-axis.

Determine a possible polynomial equation with rational coefficients that represents the roots
$2 i, \pm \sqrt{6} .$

$x^{3} - 2 i x^{2} - 6 x + 12 i$
$x^{4} - 2 x^{2} - 24$
$x^{4} - 10 x^{2} + 24$
$x^{4} - 2 x^{2} + 24$

Missing root: –2i

$x = - 2 i, 2 i, - \sqrt{6, \sqrt{6}} (x - (- 2 i)) (x - 2 i) (x - (- \sqrt{6})) (x - \sqrt{6}) = 0 (x + 2 i) (x - 2 i) (x + \sqrt{6}) (x - \sqrt{6}) = 0 (x^{2} - 4 i^{2}) (x^{2} - 6) = 0 (x^{2} - 4 (- 1)) (x^{2} - 6) = 0 (x^{2} + 4) (x^{2} - 6) = 0 x^{4} - 6 x^{2} + 4 x^{2} - 24 = 0 x^{4} - 2 x^{2} - 24 = 0$

Note

The coefficients are not all rational numbers because the missing root was not used to find the equation.

C–D) The terms were not correctly distributed resulting in a positive constant.

The process of completing the square can be used to write the equations for a ____ in standard form.

circle
ellipse
parabola
rational

Note

A rational equation uses long division to write the equation in standard form.

Problem	1	2	3	4	5	6	7	8	9	10	11	12
Origin	L27	L27	L28	L28	L26	L24	L25	L23	L28	L26	L23	L28

L = Lesson in this level, A1 = Algebra 1: Principles of Secondary Mathematics