Targeted Review Solutions

Evaluate: $[j \circ k] (- 2)$

$j (x) = {(x - 2)}^{3} - 4 k (x) = - 3 x + 2$

$j [k (- 2)] = j (8) = {(8 - 2)}^{3} - 4 = 212$

212

Find f(–2) when $f (x) = a x^{2} - 5 and f (3) = - 23 .$

$\begin{array}{rcl} - 23 & = & a {(3)}^{2} - 5 \\ - 23 & = & 9 a - 5 \\ - 18 & = & 9 a \\ a & = & - 2 \\ f (- 2) & = & - 2 {(- 2)}^{2} - 5 \\ = & - 2 (4) - 5 \\ = & - 8 - 5 \end{array}$

$f (- 2) = - 13$

Determine if the graph has a minimum or maximum point. Name the point.

minimum

$(1, - 4)$

Name the x-intercepts from the graph in the previous problem.

$(- 1.5, 0), (3.5, 0)$

Solve: $3 n^{3} + 10 n = 17 n^{2}$

$3 n^{3} - 17 n^{2} + 10 n = 0 n (3 n^{2} - 17 n + 10) = 0 n (3 n - 2) (n - 5) = 0$

$n = 0, \frac{2}{3}, 5$

Solve under the set of complex numbers: $x^{4} = 21 x^{2} + 100$

$x^{4} - 21 x^{2} - 100 = 0 (x^{2} - 25) (x^{2} + 4) = 0 (x - 5) (x + 5) (x - 2 i) (x + 2 i) = 0$

$x = \pm 5, \pm 2 i$

Use the quadratic formula to solve: $5 x^{2} - 7 x + 6 = 0$

$\begin{array}{rcl} a & = & 5, b = - 7, c = 6 \\ x & = & \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} = \frac{- (- 7) \pm \sqrt{{(- 7)}^{2} - 4 (5) (6)}}{2 (5)} \\ x & = & \frac{7 \pm \sqrt{49 - 120}}{10} = \frac{7 \pm \sqrt{- 71}}{10} \end{array}$

$x = \frac{7 \pm i \sqrt{71}}{10}$

Reflect $f (x) = \sqrt{x + 2} - 3$ over the x-axis, and then translate the function five units right, and two units up.

$a = - 1 h = - 2 + 5 = 3 k = - 3 + 2 = - 1$

$f (x) = - \sqrt{x - 3} - 1$

Note

The radicand is $x - 3$ because the equation is in the form: $f (x) = a \sqrt{x - h} + k, x \geq h$

Multiple Choice

Simplify the polynomial expression: ${(x - 3)}^{2} - 2 {(x - 1)}^{2}$

$- x^{2} + 7$
$- x^{2} - 8 x + 10$
$3 x^{2} - 10 x + 11$
$- x^{2} - 2 x + 7$

$(x - 3) (x - 3) - 2 (x - 1) (x - 1) x^{2} - 6 x + 9 - 2 (x^{2} - 2 x + 1) x^{2} - 6 x + 9 - 2 x^{2} + 4 x - 2 - x^{2} - 2 x + 7$

Note

This option is the result when the middle term is missing when the binomials are squared. 
In this option, –2 was not distributed across the second trinomial. 
This option results when the subtraction sign between the two expressions is treated as an equal sign.

Find $[f \circ g] (x)$ when $f (x) = 3 x^{2} - x - 1$ and $g (x) = - x + 6.$

x
$3 x^{2} - 35 x + 101$
$- 3 x^{2} + 8$
$3 x^{2} + x + 101$

$[f \circ g] (x) = f [g (x)] [f \circ g] (x) = 3 {(- x + 6)}^{2} - (- x + 6) - 1 [f \circ g] (x) = 3 (x^{2} - 12 x + 36) + x - 6 - 1 [f \circ g] (x) = 3 x^{2} - 36 x + 108 + x - 7 [f \circ g] (x) = 3 x^{2} - 35 x + 101$

Note

This option is [g(g(x))].
This option is [g(f(x))].
This option does not square the binomial correctly before combining terms.

Evaluate $(g - f) (4)$ when $f (x) = 3 x^{2} - 5, g (x) = x - 2 .$

–41
41
$\frac{2}{43}$
45

$\begin{array}{rcl} (g - f) (x) & = & x - 2 - (3 x^{2} - 5) \\ = & x - 2 - 3 x^{2} + 5 \\ = & - 3 x^{2} + x + 3 \\ (g - f) (4) & = & - 3 {(4)}^{2} + (4) + 3 \\ (g - f) (4) & = & - 41 \end{array}$

Note

This option is $(f - g) (4)$ .
This option is $g (4)$ divided by $f (4)$ .
This option is $(f + g) (4)$

Select all that apply.
A polynomial expression cannot contain:

fractional exponents
variables inside absolute value bars
negative coefficients
variables in the denominator

Note

Polynomial expressions can have negative coefficients, but not negative exponents.

Problem	1	2	3	4	5	6	7	8	9	10	11	12
Origin	L32	L31	L27	L27	L23	L25	L25	L18	L03	L32	L31	L03

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