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precalculus third edition pdf files should be printed before use lesson…

Question

precalculus third edition
pdf files should be printed before use
lesson (eng) lección (esp)
2 - 34. given the graph of (y = f(x)), sketch each of the following transformations. homework help
a. (y = f(x - 1)) b. (y=-f(x))
2 - 35. simplify each of the following radical expressions. homework help
a. (sqrt{18}+sqrt{32}) b. (sqrt{8y}+sqrt{18y}) c. (sqrt{75}cdotsqrt{32}) d. (sqrt{x^{7}y^{3}}cdotsqrt{x^{5}y^{8}})
2 - 36. factor each expression. homework help
a. (2x^{2}+14x - 16) b. (2x^{3}-128x)

Explanation:

Response
2 - 34
a. $y = f(x - 1)$

Step1: Recall horizontal - shift rule

The graph of $y=f(x - h)$ is a horizontal shift of the graph of $y = f(x)$. When $h>0$, the shift is to the right. Here $h = 1$, so the graph of $y=f(x)$ is shifted 1 unit to the right.

b. $y=-f(x)$

Step1: Recall reflection rule

The graph of $y=-f(x)$ is a reflection of the graph of $y = f(x)$ about the $x$-axis. For every point $(x,y)$ on the graph of $y = f(x)$, the corresponding point on the graph of $y=-f(x)$ is $(x, - y)$.

2 - 35
a. $\sqrt{18}+\sqrt{32}$

Step1: Simplify each square - root

Simplify $\sqrt{18}=\sqrt{9\times2}=3\sqrt{2}$ and $\sqrt{32}=\sqrt{16\times2}=4\sqrt{2}$.

Step2: Combine like terms

$3\sqrt{2}+4\sqrt{2}=(3 + 4)\sqrt{2}=7\sqrt{2}$

b. $\sqrt{8y}+\sqrt{18y}$

Step1: Simplify each square - root

$\sqrt{8y}=\sqrt{4\times2y}=2\sqrt{2y}$ and $\sqrt{18y}=\sqrt{9\times2y}=3\sqrt{2y}$.

Step2: Combine like terms

$2\sqrt{2y}+3\sqrt{2y}=(2 + 3)\sqrt{2y}=5\sqrt{2y}$

c. $\sqrt{75}\cdot\sqrt{32}$

Step1: Simplify each square - root

$\sqrt{75}=\sqrt{25\times3}=5\sqrt{3}$ and $\sqrt{32}=\sqrt{16\times2}=4\sqrt{2}$.

Step2: Multiply the simplified square - roots

$5\sqrt{3}\times4\sqrt{2}=(5\times4)\sqrt{3\times2}=20\sqrt{6}$

d. $\sqrt{x^{7}y^{3}}\cdot\sqrt{x^{5}y^{8}}$

Step1: Use the product rule of square - roots $\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}$

$\sqrt{x^{7}y^{3}\cdot x^{5}y^{8}}=\sqrt{x^{7 + 5}y^{3+8}}=\sqrt{x^{12}y^{11}}$.

Step2: Simplify the square - root

$\sqrt{x^{12}y^{11}}=x^{6}y^{5}\sqrt{y}$

2 - 36
a. $2x^{2}+14x - 16$

Step1: Factor out the greatest common factor (GCF)

The GCF of $2x^{2},14x$ and $-16$ is 2. So $2x^{2}+14x - 16=2(x^{2}+7x - 8)$.

Step2: Factor the quadratic expression inside the parentheses

We need to find two numbers that multiply to $-8$ and add up to 7. The numbers are 8 and - 1. So $x^{2}+7x - 8=(x + 8)(x-1)$. Then $2x^{2}+14x - 16=2(x + 8)(x - 1)$

b. $2x^{3}-128x$

Step1: Factor out the GCF

The GCF of $2x^{3}$ and $-128x$ is $2x$. So $2x^{3}-128x=2x(x^{2}-64)$.

Step2: Use the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$

Here $a=x$ and $b = 8$, so $x^{2}-64=(x + 8)(x - 8)$. Then $2x^{3}-128x=2x(x + 8)(x - 8)$

Answer:

Step1: Factor out the greatest common factor (GCF)

The GCF of $2x^{2},14x$ and $-16$ is 2. So $2x^{2}+14x - 16=2(x^{2}+7x - 8)$.

Step2: Factor the quadratic expression inside the parentheses

We need to find two numbers that multiply to $-8$ and add up to 7. The numbers are 8 and - 1. So $x^{2}+7x - 8=(x + 8)(x-1)$. Then $2x^{2}+14x - 16=2(x + 8)(x - 1)$

b. $2x^{3}-128x$

Step1: Factor out the GCF

The GCF of $2x^{3}$ and $-128x$ is $2x$. So $2x^{3}-128x=2x(x^{2}-64)$.

Step2: Use the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$

Here $a=x$ and $b = 8$, so $x^{2}-64=(x + 8)(x - 8)$. Then $2x^{3}-128x=2x(x + 8)(x - 8)$