Solved on Jan 24, 2024

Draw graphs for $y=\frac{1}{2}|x-3|-5$ and $y=-2^{(x+1)}+4$ , identifying all key points and asymptotes.

STEP 1

Assumptions
1. We are drawing the graph of two functions.
2. The first function is $y=\frac{1}{2}|x-3|-5$ .
3. The second function is $y=-2^{(x+1)}+4$ .
4. We will identify key points such as intercepts and asymptotes.
5. The graphs will be drawn on a standard Cartesian plane.

STEP 2

To graph the first function $y=\frac{1}{2}|x-3|-5$ , we will consider the absolute value function $|x-3|$ and apply transformations to it.

STEP 3

The graph of $y=|x-3|$ is a V-shaped graph with its vertex at the point (3,0). This is because the absolute value function reflects all negative inputs to positive outputs, creating a V shape.

STEP 4

The transformation $\frac{1}{2}|x-3|$ compresses the graph of $|x-3|$ vertically by a factor of $\frac{1}{2}$ . This means that all y-values will be halved.

STEP 5

The transformation $\frac{1}{2}|x-3|-5$ then shifts the graph of $\frac{1}{2}|x-3|$ downward by 5 units. This means that all y-values will be reduced by 5.

STEP 6

To find the x-intercept(s), we set $y=0$ and solve for $x$ :
$0 = \frac{1}{2}|x-3|-5$

STEP 7

Add 5 to both sides of the equation:
$5 = \frac{1}{2}|x-3|$

STEP 8

Multiply both sides by 2 to get rid of the fraction:
$10 = |x-3|$

STEP 9

Solve the absolute value equation, which has two solutions:
$x-3 = 10 \quad \text{or} \quad x-3 = -10$

STEP 10

Solve for $x$ in both cases:
$x = 13 \quad \text{or} \quad x = -7$

STEP 11

The x-intercepts are at (13,0) and (-7,0).

STEP 12

The y-intercept occurs when $x=0$ . Plug in $x=0$ into the original equation to find the y-intercept:
$y = \frac{1}{2}|0-3|-5$

STEP 13

Simplify the absolute value and the arithmetic:
$y = \frac{1}{2}(3) - 5 = \frac{3}{2} - 5 = -\frac{7}{2}$

STEP 14

The y-intercept is at (0, -3.5).

STEP 15

There are no horizontal or vertical asymptotes for the graph of $y=\frac{1}{2}|x-3|-5$ .

STEP 16

Now, let's graph the second function $y=-2^{(x+1)}+4$ .

STEP 17

The base graph $y=2^x$ is an exponential growth function with a horizontal asymptote at $y=0$ .

STEP 18

The transformation $2^{(x+1)}$ shifts the graph of $y=2^x$ to the left by 1 unit.

STEP 19

The transformation $-2^{(x+1)}$ reflects the graph of $y=2^{(x+1)}$ across the x-axis, turning the exponential growth into exponential decay.

STEP 20

The transformation $-2^{(x+1)}+4$ then shifts the graph of $y=-2^{(x+1)}$ upward by 4 units.

STEP 21

The horizontal asymptote is affected by the vertical shift and is now at $y=4$ .

STEP 22

To find the x-intercept(s), set $y=0$ and solve for $x$ :
$0 = -2^{(x+1)}+4$

STEP 23

Subtract 4 from both sides:
$-4 = -2^{(x+1)}$

STEP 24

Divide by -1:
$4 = 2^{(x+1)}$

STEP 25

Recognize that $4 = 2^2$ and solve for $x$ :
$2^2 = 2^{(x+1)}$

STEP 26

Since the bases are the same, the exponents must be equal:
$2 = x+1$

STEP 27

Solve for $x$ :
$x = 1$

STEP 28

The x-intercept is at (1,0).

STEP 29

The y-intercept occurs when $x=0$ . Plug in $x=0$ into the original equation to find the y-intercept:
$y = -2^{(0+1)}+4$

STEP 30

Simplify the exponent and the arithmetic:
$y = -2^1 + 4 = -2 + 4 = 2$
The y-intercept is at (0, 2).
The key points and asymptotes for the graphs are as follows:
For $y=\frac{1}{2}|x-3|-5$ : - x-intercepts: (13,0) and (-7,0) - y-intercept: (0, -3.5) - No asymptotes
For $y=-2^{(x+1)}+4$ : - x-intercept: (1,0) - y-intercept: (0, 2) - Horizontal asymptote: $y=4$

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Draw graphs for y=12∣x−3∣−5y=\frac{1}{2}|x-3|-5y=21​∣x−3∣−5 and y=−2(x+1)+4y=-2^{(x+1)}+4y=−2(x+1)+4, identifying all key points and asymptotes.

STEP 1

STEP 2

To graph the first function y=12∣x−3∣−5y=\frac{1}{2}|x-3|-5y=21​∣x−3∣−5, we will consider the absolute value function ∣x−3∣|x-3|∣x−3∣ and apply transformations to it.

STEP 3

The graph of y=∣x−3∣y=|x-3|y=∣x−3∣ is a V-shaped graph with its vertex at the point (3,0). This is because the absolute value function reflects all negative inputs to positive outputs, creating a V shape.

STEP 4

The transformation 12∣x−3∣\frac{1}{2}|x-3|21​∣x−3∣ compresses the graph of ∣x−3∣|x-3|∣x−3∣ vertically by a factor of 12\frac{1}{2}21​. This means that all y-values will be halved.

STEP 5

The transformation 12∣x−3∣−5\frac{1}{2}|x-3|-521​∣x−3∣−5 then shifts the graph of 12∣x−3∣\frac{1}{2}|x-3|21​∣x−3∣ downward by 5 units. This means that all y-values will be reduced by 5.

STEP 6

To find the x-intercept(s), we set y=0y=0y=0 and solve for xxx:0=12∣x−3∣−50 = \frac{1}{2}|x-3|-50=21​∣x−3∣−5

STEP 7

Add 5 to both sides of the equation:5=12∣x−3∣5 = \frac{1}{2}|x-3|5=21​∣x−3∣

STEP 8

Multiply both sides by 2 to get rid of the fraction:10=∣x−3∣10 = |x-3|10=∣x−3∣

STEP 9

Solve the absolute value equation, which has two solutions:x−3=10orx−3=−10x-3 = 10 \quad \text{or} \quad x-3 = -10x−3=10orx−3=−10

STEP 10

Solve for xxx in both cases:x=13orx=−7x = 13 \quad \text{or} \quad x = -7x=13orx=−7

STEP 11

The x-intercepts are at (13,0) and (-7,0).

STEP 12

The y-intercept occurs when x=0x=0x=0. Plug in x=0x=0x=0 into the original equation to find the y-intercept:y=12∣0−3∣−5y = \frac{1}{2}|0-3|-5y=21​∣0−3∣−5

STEP 13

Simplify the absolute value and the arithmetic:y=12(3)−5=32−5=−72y = \frac{1}{2}(3) - 5 = \frac{3}{2} - 5 = -\frac{7}{2}y=21​(3)−5=23​−5=−27​

STEP 14

The y-intercept is at (0, -3.5).

STEP 15

There are no horizontal or vertical asymptotes for the graph of y=12∣x−3∣−5y=\frac{1}{2}|x-3|-5y=21​∣x−3∣−5.

STEP 16

Now, let's graph the second function y=−2(x+1)+4y=-2^{(x+1)}+4y=−2(x+1)+4.

STEP 17

The base graph y=2xy=2^xy=2x is an exponential growth function with a horizontal asymptote at y=0y=0y=0.

STEP 18

The transformation 2(x+1)2^{(x+1)}2(x+1) shifts the graph of y=2xy=2^xy=2x to the left by 1 unit.

STEP 19

The transformation −2(x+1)-2^{(x+1)}−2(x+1) reflects the graph of y=2(x+1)y=2^{(x+1)}y=2(x+1) across the x-axis, turning the exponential growth into exponential decay.

STEP 20

The transformation −2(x+1)+4-2^{(x+1)}+4−2(x+1)+4 then shifts the graph of y=−2(x+1)y=-2^{(x+1)}y=−2(x+1) upward by 4 units.

STEP 21

The horizontal asymptote is affected by the vertical shift and is now at y=4y=4y=4.

STEP 22

To find the x-intercept(s), set y=0y=0y=0 and solve for xxx:0=−2(x+1)+40 = -2^{(x+1)}+40=−2(x+1)+4

STEP 23

Subtract 4 from both sides:−4=−2(x+1)-4 = -2^{(x+1)}−4=−2(x+1)

STEP 24

Divide by -1:4=2(x+1)4 = 2^{(x+1)}4=2(x+1)

STEP 25

Recognize that 4=224 = 2^24=22 and solve for xxx:22=2(x+1)2^2 = 2^{(x+1)}22=2(x+1)

STEP 26

Since the bases are the same, the exponents must be equal:2=x+12 = x+12=x+1

STEP 27

Solve for xxx:x=1x = 1x=1

STEP 28

The x-intercept is at (1,0).

STEP 29

The y-intercept occurs when x=0x=0x=0. Plug in x=0x=0x=0 into the original equation to find the y-intercept:y=−2(0+1)+4y = -2^{(0+1)}+4y=−2(0+1)+4

STEP 30

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Draw graphs for $y=\frac{1}{2}|x-3|-5$ and $y=-2^{(x+1)}+4$ , identifying all key points and asymptotes.

To graph the first function $y=\frac{1}{2}|x-3|-5$ , we will consider the absolute value function $|x-3|$ and apply transformations to it.

The graph of $y=|x-3|$ is a V-shaped graph with its vertex at the point (3,0). This is because the absolute value function reflects all negative inputs to positive outputs, creating a V shape.

The transformation $\frac{1}{2}|x-3|$ compresses the graph of $|x-3|$ vertically by a factor of $\frac{1}{2}$ . This means that all y-values will be halved.

The transformation $\frac{1}{2}|x-3|-5$ then shifts the graph of $\frac{1}{2}|x-3|$ downward by 5 units. This means that all y-values will be reduced by 5.

To find the x-intercept(s), we set $y=0$ and solve for $x$ :
$0 = \frac{1}{2}|x-3|-5$

Add 5 to both sides of the equation:
$5 = \frac{1}{2}|x-3|$

Multiply both sides by 2 to get rid of the fraction:
$10 = |x-3|$

Solve the absolute value equation, which has two solutions:
$x-3 = 10 \quad \text{or} \quad x-3 = -10$

Solve for $x$ in both cases:
$x = 13 \quad \text{or} \quad x = -7$

The y-intercept occurs when $x=0$ . Plug in $x=0$ into the original equation to find the y-intercept:
$y = \frac{1}{2}|0-3|-5$

Simplify the absolute value and the arithmetic:
$y = \frac{1}{2}(3) - 5 = \frac{3}{2} - 5 = -\frac{7}{2}$

There are no horizontal or vertical asymptotes for the graph of $y=\frac{1}{2}|x-3|-5$ .

Now, let's graph the second function $y=-2^{(x+1)}+4$ .

The base graph $y=2^x$ is an exponential growth function with a horizontal asymptote at $y=0$ .

The transformation $2^{(x+1)}$ shifts the graph of $y=2^x$ to the left by 1 unit.

The transformation $-2^{(x+1)}$ reflects the graph of $y=2^{(x+1)}$ across the x-axis, turning the exponential growth into exponential decay.

The transformation $-2^{(x+1)}+4$ then shifts the graph of $y=-2^{(x+1)}$ upward by 4 units.

The horizontal asymptote is affected by the vertical shift and is now at $y=4$ .

To find the x-intercept(s), set $y=0$ and solve for $x$ :
$0 = -2^{(x+1)}+4$

Subtract 4 from both sides:
$-4 = -2^{(x+1)}$

Divide by -1:
$4 = 2^{(x+1)}$

Recognize that $4 = 2^2$ and solve for $x$ :
$2^2 = 2^{(x+1)}$

Since the bases are the same, the exponents must be equal:
$2 = x+1$

Solve for $x$ :
$x = 1$

The y-intercept occurs when $x=0$ . Plug in $x=0$ into the original equation to find the y-intercept:
$y = -2^{(0+1)}+4$