Targeted Review Solutions

For problems 1–4, use the graphs of f(x) and g(x).

$(f + g) (- 4)$

$\begin{array}{rcl} (f + g) (- 4) & = & f (- 4) + g (- 4) \\ = & 0 + 3 \end{array}$

$(f \circ g) (2)$

$\begin{array}{rcl} (f \circ g) (2) & = & f (g (2)) \\ = & f (3) \\ (f \circ g) (2) & = & - 2 \end{array}$

–2

$(g \circ g) (6)$

$\begin{array}{rcl} (g \circ g) (6) & = & g (g (6)) \\ = & g (5) \\ g (x) & = & \frac{1}{2} x + 2 \\ g (5) & = & \frac{1}{2} (5) + 2 \\ g (5) & = & \frac{9}{2} \end{array}$  

$\begin{array}{rcl} \frac{9}{2} \end{array}$

$(\frac{f}{g}) (- 1)$

$\begin{array}{rcl} (\frac{f}{g}) (- 1) & = & \frac{f (- 1)}{g (- 1)} \\ = & \frac{0}{1.5} \end{array}$

Determine all possible rational roots for $f (x) = 3 x^{3} + 2 x^{2} - 19 x + 6$ .

$RRT = \frac{\pm 1, \pm 2, \pm 3, \pm 6}{\pm 1, \pm 3}$

$RRT = = \pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{3}, \pm \frac{2}{3}$

Write the polynomial from problem 5 as a product of its rational factors.

$3 x^{2} + 8 x - 3 = 0 (3 x - 1) (x + 3) = 0 x = \frac{1}{3}, - 3 x = - 3, \frac{1}{3}, 2$

$(x + 3) (3 x - 1) (x - 2) = 0$

Identify the end behavior from the sketch of the function:

a > 0 

$x \to \pm \infty, f (x) \to + \infty$

State the roots and the multiplicities for $b (x) = x^{2} {(x + 7)}^{3} (x - 6)$ .

$0 = x^{2} {(x + 7)}^{3} (x - 6) x^{2} = 0 x + 7 = 0 x - 6 = 0 x = 0, - 7, 6$  

When $x = 0$ , the multiplicity is two. When $x = - 7$ , the multiplicity is three. When $x = 6$ , the multiplicity is one.

Multiple Choice

Determine the end behavior for $h (x) = x {(x - 3)}^{2} (x + 1) (x - 5) .$

$x \to \pm \infty, f (x) \to - \infty$
$x \to \pm \infty, f (x) \to + \infty$
$x \to - \infty, f (x) \to - \infty,$ and $x \to + \infty, f (x) \to + \infty$
$x \to - \infty, f (x) \to + \infty,$ and $x \to + \infty, f (x) \to - \infty$

n = 1 + 2 + 1 + 1 = 5

a > 0

Note

A, B) These options represent end behavior for an even degree function.

A, D) These options represent a function where $a < 0$ .

Determine $g \circ f$ when $f (x) = 3 x + 1,$ and $g (x) = \frac{- x + 1}{2} .$

$- \frac{3}{2} x + \frac{5}{2}$
$- \frac{3}{2} x$
$\frac{3}{2} x + \frac{5}{2}$
$\frac{3}{2} x + 1$

$\begin{array}{rcl} g \circ f & = & g (f (x)) \\ = & \frac{- (3 x + 1) + 1}{2} \\ = & \frac{- 3 x - 1 + 1}{2} \\ g \circ f & = & - \frac{3 x}{2} \end{array}$

Note

A) This option is f(g(x)).

C, D) These options ignore the negative symbol in g(x).

Determine q(3) for the function q(x), when $q (- 2) = - \frac{1}{3} .$
$q (x) = a x^{2} - 3 a$

–2
$- \frac{1}{3}$
2
3

$\begin{array}{rcl} \begin{matrix} q (x) = a x^{2} - 3 a \\ q (x) = a (x^{2} - 3) \\ q (- 2) = - \frac{1}{3} & q (x) = - \frac{1}{3} (x^{2} - 3) \\ - \frac{1}{3} = a ({(- 2)}^{2} - 3) & q (3) = - \frac{1}{3} (3^{2} - 3) \\ - \frac{1}{3} = a (4 - 3) & q (3) = - \frac{1}{3} (6) \\ a = - \frac{1}{3} & q (3) = - 2 \end{matrix} \end{array}$

Note

This option is the value of a, not q(3). 
The correct answer is negative, not positive. 
This option is the given x-value.

Select the function that contains a double root at f(5) = 0.

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

Double root is a multiplicity of 2.
There is a zero at $(5, 0) .$

Note

A, C) These options have a double root at –5. 

This option has a double root at 0.

Problem	1	2	3	4	5	6	7	8	9	10	11	12
Origin	L31	L32	L32	L31	L34	L34	L33	L34	L33	L32	L31	L34

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