Question

review & refresh
write a linear function that relates y to x.
1.
2.

x	0	1.5	3	4.5
y	5	4	3	2

the vertices of a figure are given. draw the figure and its image after a dilation with the given scale factor. identify the type of dilation.

1. $a(-3, 1), b(-1, 3), c(-1, 1); k=3$
2. $j(2, 4), k(6, 10), l(8, 10), m(8, 4); k=\frac{1}{4}$

concepts, skills, & problem solving
comparing functions graph each equation. decide whether each graph represents a linear or nonlinear function. (see exploration 1, p. 295.)

1. $h = 5 + 6t$ equation 1

$h = 5 + 6t^2$ equation 2

1. $y = -\frac{x}{3}$ equation 1

$y = -\frac{3}{x}$ equation 2

identifying functions from tables does the table represent a linear or nonlinear function? explain.
7.

x	0	1	2	3
y	4	8	12	16

8.

x	6	5	4	3
y	21	15	10	6

identifying functions from equations does the equation represent a linear or nonlinear function? explain.

1. $2x + 3y = 7$
2. $y + x = 4x + 5$
3. $y = \frac{8}{x^2}$

identifying functions from graphs does the graph represent a linear or nonlinear function? explain.
12.
13.

Explanation:

Step1: Find slope from graph

Choose two points, e.g., $(2,0)$ and $(0,-1)$. Slope $m=\frac{0-(-1)}{2-0}=\frac{1}{2}$

Step2: Identify y-intercept

The line crosses y-axis at $(0,-1)$, so $b=-1$.

Step3: Write linear function

Substitute $m$ and $b$ into $y=mx+b$.
$y=\frac{1}{2}x - 1$

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Step1: Calculate slope from table

Use points $(0,5)$ and $(1.5,4)$. $m=\frac{4-5}{1.5-0}=-\frac{2}{3}$

Step2: Identify y-intercept

When $x=0$, $y=5$, so $b=5$.

Step3: Write linear function

Substitute $m$ and $b$ into $y=mx+b$.
$y=-\frac{2}{3}x + 5$

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Step1: Dilate each vertex by $k=3$

$A'(-3\times3, 1\times3)=(-9,3)$; $B'(-1\times3, 3\times3)=(-3,9)$; $C'(-1\times3,1\times3)=(-3,3)$

Step2: Identify dilation type

Scale factor $k>1$, so it is an enlargement.

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Step1: Dilate each vertex by $k=\frac{1}{4}$

$J'(2\times\frac{1}{4}, 4\times\frac{1}{4})=(0.5,1)$; $K'(6\times\frac{1}{4},10\times\frac{1}{4})=(1.5,2.5)$; $L'(8\times\frac{1}{4},10\times\frac{1}{4})=(2,2.5)$; $M'(8\times\frac{1}{4},4\times\frac{1}{4})=(2,1)$

Step2: Identify dilation type

Scale factor $0

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Step1: Analyze Equation 1: $h=5+6t$

It follows $y=mx+b$ form, so linear. Graph is a straight line.

Step2: Analyze Equation 2: $h=5+6t^2$

Has squared variable $t^2$, so nonlinear. Graph is a parabola.

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Step1: Analyze Equation 1: $y=-\frac{x}{3}$

It follows $y=mx+b$ form, so linear. Graph is a straight line.

Step2: Analyze Equation 2: $y=-\frac{3}{x}$

Has variable in denominator, so nonlinear. Graph is a hyperbola.

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Step1: Check rate of change

$\frac{8-4}{1-0}=4$, $\frac{12-8}{2-1}=4$, $\frac{16-12}{3-2}=4$

Step2: Classify function

Constant rate of change, so linear.

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Step1: Check rate of change

$\frac{15-21}{5-6}=6$, $\frac{10-15}{4-5}=5$, $\frac{6-10}{3-4}=4$

Step2: Classify function

Non-constant rate of change, so nonlinear.

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Step1: Rewrite in standard linear form

$2x+3y=7$ can be rewritten as $y=-\frac{2}{3}x+\frac{7}{3}$, follows $y=mx+b$.

Step2: Classify function

Linear, no exponents >1 on variables.

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Step1: Simplify the equation

$y+x=4x+5$ simplifies to $y=3x+5$, follows $y=mx+b$.

Step2: Classify function

Linear, no exponents >1 on variables.

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Step1: Analyze the equation $y=\frac{8}{x^2}$

Variable $x$ is squared and in denominator, does not follow linear form.

Step2: Classify function

Nonlinear, has variable with negative exponent.

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Step1: Analyze the graph

Graph is a straight line with constant slope.

Step2: Classify function

Linear, straight line graph represents linear function.

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Step1: Analyze the graph

Graph is a curve with changing slope, not a straight line.

Step2: Classify function

Nonlinear, curved graph represents nonlinear function.

Answer:

1. $\boldsymbol{y=\frac{1}{2}x - 1}$
2. $\boldsymbol{y=-\frac{2}{3}x + 5}$
3. Dilation image vertices: $A'(-9,3)$, $B'(-3,9)$, $C'(-3,3)$; Type: Enlargement
4. Dilation image vertices: $J'(0.5,1)$, $K'(1.5,2.5)$, $L'(2,2.5)$, $M'(2,1)$; Type: Reduction
5. Equation 1: Linear (straight line graph); Equation 2: Nonlinear (parabola graph)
6. Equation 1: Linear (straight line graph); Equation 2: Nonlinear (hyperbola graph)
7. Linear; constant rate of change of 4 between $x$ and $y$
8. Nonlinear; rate of change between $x$ and $y$ is not constant
9. Linear; can be rewritten as $y=-\frac{2}{3}x+\frac{7}{3}$, fits linear form $y=mx+b$
10. Linear; simplifies to $y=3x+5$, fits linear form $y=mx+b$
11. Nonlinear; variable $x$ is in the denominator (equivalent to $x^{-2}$), does not fit linear form
12. Linear; the graph is a straight line with a constant slope
13. Nonlinear; the graph is a curve with a changing slope