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

Studdy Solution

STEP 1

Assumptions
1. We are drawing the graph of two functions.
2. The first function is y=12∣x−3∣−5y=\frac{1}{2}|x-3|-5.
3. The second function is y=−2(x+1)+4y=-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=12∣x−3∣−5y=\frac{1}{2}|x-3|-5, we will consider the absolute value function ∣x−3∣|x-3| and apply transformations to it.

STEP 3

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

STEP 5

The transformation 12∣x−3∣−5\frac{1}{2}|x-3|-5 then shifts the graph of 12∣x−3∣\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=0y=0 and solve for xx:
0=12∣x−3∣−50 = \frac{1}{2}|x-3|-5

STEP 7

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

STEP 8

Multiply both sides by 2 to get rid of the fraction:
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 = -10

STEP 10

Solve for xx in both cases:
x=13orx=−7x = 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=0x=0. Plug in x=0x=0 into the original equation to find the y-intercept:
y=12∣0−3∣−5y = \frac{1}{2}|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}

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|-5.

STEP 16

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

STEP 17

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

STEP 18

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

STEP 19

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

STEP 22

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

STEP 23

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

STEP 24

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

STEP 25

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

STEP 26

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

STEP 27

Solve for xx:
x=1x = 1

STEP 28

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

STEP 29

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

STEP 30

Simplify the exponent and the arithmetic:
y=−21+4=−2+4=2y = -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=12∣x−3∣−5y=\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)+4y=-2^{(x+1)}+4: - x-intercept: (1,0) - y-intercept: (0, 2) - Horizontal asymptote: y=4y=4

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