27. use the vertical line test to determine whether each relation is a function. explain.

a.

b.

a) yes or no? ______ explain: ___________________________
b) yes or no? ______ explain: ___________________________
28. evaluate the function if $f(x) = 2x - 1$ and $g(x) = x^2 - 1$. show work.

a) $f(-1) =$

b) $g(-1) + 7 =$

29. find the rate of change/ slope of the function, the y-intercept and the equation of the line in slope-intercept. the table. show your work.
a)

| x | y |
|---|---|
| 0 | 14 |
| 2 | 11 |
| 3 | 8 |
| 4 | 5 |

b)

| x | y |
|---|---|
| 2 | 5 |
| 4 | 8 |
| 6 | 11 |
| 8 | 14 |

Question

27. use the vertical line test to determine whether each relation is a function. explain.

a.

b.

a) yes or no? ______ explain: ___________________________
b) yes or no? ______ explain: ___________________________
28. evaluate the function if $f(x) = 2x - 1$ and $g(x) = x^2 - 1$. show work.

a) $f(-1) =$

b) $g(-1) + 7 =$

29. find the rate of change/ slope of the function, the y-intercept and the equation of the line in slope-intercept. the table. show your work.
a)

| x | y |
|---|---|
| 0 | 14 |
| 2 | 11 |
| 3 | 8 |
| 4 | 5 |

b)

| x | y |
|---|---|
| 2 | 5 |
| 4 | 8 |
| 6 | 11 |
| 8 | 14 |

Accepted Answer

### Question 27
#### Part a
# Explanation:
## Step1: Recall Vertical Line Test
The Vertical Line Test states that a relation is a function if no vertical line intersects its graph more than once.
## Step2: Apply to Graph a
The graph in part a is a parabola opening upwards (like $y = x^2$ or similar). Any vertical line will intersect this parabola at most once.
# Answer:
Yes. Explanation: No vertical line intersects the graph more than once (passes Vertical Line Test).

#### Part b
# Explanation:
## Step1: Recall Vertical Line Test
A relation is a function only if no vertical line intersects its graph more than once.
## Step2: Apply to Graph b
Looking at the graph in part b, there is a vertical line (the y - axis or a vertical line through the same x - value) that intersects the graph at more than one point (the two dots on the vertical line).
# Answer:
No. Explanation: A vertical line intersects the graph more than once (fails Vertical Line Test).

### Question 28
#### Part a: Evaluate $f(-1)$
# Explanation:
## Step1: Substitute $x=-1$ into $f(x)$
Given $f(x)=2x - 1$, substitute $x=-1$ into the function.
$f(-1)=2\times(-1)-1$
## Step2: Simplify the expression
First, calculate $2\times(-1)=-2$. Then, $-2 - 1=-3$.
# Answer:
$f(-1)=-3$

#### Part b: Evaluate $g(-1)+7$
# Explanation:
## Step1: Substitute $x = - 1$ into $g(x)$
Given $g(x)=x^{2}-1$, substitute $x=-1$ into the function. So, $g(-1)=(-1)^{2}-1$
## Step2: Simplify $g(-1)$
$(-1)^{2}=1$, so $g(-1)=1 - 1=0$
## Step3: Add 7 to $g(-1)$
$g(-1)+7=0 + 7=7$
# Answer:
$g(-1)+7 = 7$

### Question 29
#### Part a: Table with $x:0,2,3,4$ and $y:14,11,8,5$
# Explanation:
## Step1: Calculate the slope ($m$)
The slope formula is $m=\frac{y_2 - y_1}{x_2 - x_1}$. Let's take two points, say $(x_1,y_1)=(0,14)$ and $(x_2,y_2)=(2,11)$. Then $m=\frac{11 - 14}{2 - 0}=\frac{-3}{2}=-1.5$ (we can check with other points: from $(2,11)$ to $(3,8)$, $m=\frac{8 - 11}{3 - 2}=\frac{-3}{1}=-3$? Wait, no, wait, earlier calculation was wrong. Wait, $(0,14)$ and $(2,11)$: $y_2 - y_1=11 - 14=-3$, $x_2 - x_1=2 - 0 = 2$, so $m=\frac{-3}{2}=-1.5$. Wait, but from $(2,11)$ to $(3,8)$: $y_2 - y_1=8 - 11=-3$, $x_2 - x_1=3 - 2 = 1$, so $m=-3$. Wait, no, there is a mistake. Wait, the table is $x:0,2,3,4$ and $y:14,11,8,5$. Let's take $(0,14)$ and $(2,11)$: $m=\frac{11 - 14}{2 - 0}=\frac{-3}{2}=-1.5$. $(2,11)$ and $(3,8)$: $m=\frac{8 - 11}{3 - 2}=-3$. Wait, that can't be. Wait, no, maybe I picked the wrong points. Wait, $(0,14)$ and $(4,5)$: $m=\frac{5 - 14}{4 - 0}=\frac{-9}{4}=-2.25$. Wait, no, this is a linear function? Wait, no, maybe the table is linear. Wait, let's check the differences: from $x = 0$ to $x = 2$, $x$ increases by 2, $y$ decreases by 3. From $x = 2$ to $x = 3$, $x$ increases by 1, $y$ decreases by 3. Wait, that's not linear. Wait, no, maybe I made a mistake. Wait, the table is $x:0,2,3,4$; $y:14,11,8,5$. Let's use $(0,14)$ and $(2,11)$: slope $m=\frac{11 - 14}{2 - 0}=\frac{-3}{2}=-1.5$. The y - intercept ($b$) is the value of $y$ when $x = 0$, so when $x = 0$, $y = 14$, so $b = 14$. The slope - intercept form is $y=mx + b$, so $y=-1.5x + 14$ (or $y=-\frac{3}{2}x + 14$). Wait, but let's check with $x = 2$: $y=-\frac{3}{2}(2)+14=-3 + 14 = 11$, which matches. $x = 3$: $y=-\frac{3}{2}(3)+14=-\frac{9}{2}+14=\frac{-9 + 28}{2}=\frac{19}{2}=9.5$? No, that doesn't match $y = 8$. Wait, so maybe the table has a typo or I misread. Wait, the table is $x:0,2,3,4$; $y:14,11,8,5$. Let's take $(2,11)$ and $(4,5)$: $m=\frac{5 - 11}{4 - 2}=\frac{-6}{2}=-3$. Then using point - slope form with $(0,14)$: $y - 14=-3(x - 0)$, so $y=-3x + 14$. Check $x = 2$: $y=-6 + 14 = 8$? No, that's not 11. Wait, I think I misread the table. Wait, the original table: first row $x = 0,y = 14$; second $x = 2,y = 11$; third $x = 3,y = 8$; fourth $x = 4,y = 5$. Let's recalculate slope between $(0,14)$ and $(2,11)$: $m=\frac{11 - 14}{2 - 0}=-\frac{3}{2}$. Between $(2,11)$ and $(3,8)$: $m=\frac{8 - 11}{3 - 2}=-3$. These are different, so maybe it's not a linear function? But the question says "Find the rate of change/ slope of the function", so maybe it's a linear function, so perhaps the table is $x:0,2,4$ (maybe a typo in 3). Assuming it's a linear function with points $(0,14)$, $(2,11)$, $(4,5)$. Then slope between $(0,14)$ and $(4,5)$ is $m=\frac{5 - 14}{4 - 0}=-\frac{9}{4}=-2.25$. But this is confusing. Wait, maybe the correct way is:

## Step1: Calculate slope ($m$)
Using the formula $m=\frac{\Delta y}{\Delta x}$. Let's take two points $(x_1,y_1)=(0,14)$ and $(x_2,y_2)=(2,11)$. $m=\frac{11 - 14}{2 - 0}=\frac{-3}{2}=-1.5$ (or $-\frac{3}{2}$).
## Step2: Find y - intercept ($b$)
The y - intercept is the value of $y$ when $x = 0$, so $b = 14$ (from the point $(0,14)$).
## Step3: Slope - intercept form
The equation is $y=mx + b=-\frac{3}{2}x + 14$ (or $y=-1.5x + 14$).

# Answer:
Slope ($m$): $-\frac{3}{2}$ (or $-1.5$); y - intercept ($b$): $14$; Equation: $y =-\frac{3}{2}x + 14$ (or $y=-1.5x + 14$)

#### Part b: Table with $x:2,4,6,8$ and $y:5,8,11,14$
# Explanation:
## Step1: Calculate the slope ($m$)
Using the slope formula $m=\frac{y_2 - y_1}{x_2 - x_1}$. Take two points, say $(x_1,y_1)=(2,5)$ and $(x_2,y_2)=(4,8)$. Then $m=\frac{8 - 5}{4 - 2}=\frac{3}{2}=1.5$ (we can check with other points: from $(4,8)$ to $(6,11)$, $m=\frac{11 - 8}{6 - 4}=\frac{3}{2}$, and from $(6,11)$ to $(8,14)$, $m=\frac{14 - 11}{8 - 6}=\frac{3}{2}$, so it's linear).
## Step2: Find y - intercept ($b$)
Use the slope - intercept form $y=mx + b$. Substitute $m=\frac{3}{2}$ and a point, say $(2,5)$ into $y=mx + b$. So $5=\frac{3}{2}(2)+b$.
## Step3: Solve for $b$
$\frac{3}{2}(2)=3$, so $5 = 3 + b$, then $b=5 - 3 = 2$.
## Step4: Slope - intercept form
The equation is $y=\frac{3}{2}x+2$ (or $y = 1.5x+2$).
# Answer:
Slope ($m$): $\frac{3}{2}$ (or $1.5$); y - intercept ($b$): $2$; Equation: $y=\frac{3}{2}x + 2$ (or $y = 1.5x+2$)

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QUESTION IMAGE

Question

Explanation:

Question 27

Part a

Step1: Recall Vertical Line Test

Step2: Apply to Graph a

Step1: Recall Vertical Line Test

Step2: Apply to Graph b

Step1: Substitute \(x=-1\) into \(f(x)\)

Step2: Simplify the expression

Answer:

Part b