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ser-zykov [4K]
2 years ago
15

What is the value of g( − 3) when g(x) = 2x-2

Mathematics
2 answers:
AnnyKZ [126]2 years ago
3 0

Hello,

g(x) = 2x - 2

g(-3) = 2 \times (- 3) - 2 = -6 - 2 = -8

FromTheMoon [43]2 years ago
3 0

Answer: g(-3) = -8

Given that:

  • g(x) = 2x - 2

To find g(-3), insert (x = -3):

⇒ g(-3) = 2(-3) - 2

⇒ g(-3) = -6 - 2

⇒ g(-3) = -8

Find more at: brainly.com/question/926984

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An x-intercept is the point where the function passes the x axis at y=0.
The y-intercept is the point where the function crosses the y axis at x=0

1. It is an x-intercept, so y = 0. The ordered pair would be (-6, 0)
2. It is a y-intercept, so x = 0. The ordered pair would be (0, -2.3)
3. It is a y-intercept, so x=0. The ordered pair would be (0, 3/4)

To find the x-intercept, set y = 0. To find the y-intercept, set x=0

4. y-intercept: y = 3(0) -9. y = -9 The y-intercept is at (0, -9)
x-intercept: 0 = 3x -9. 9 = 3x. x = 3. The x-intercept is at (3, 0)

5. x intercept: 0 = 5x +10. -10 = 5x. x = -2. The x-intercept is at (-2, 0)
y-intercept: y = 5(0) + 10. y = 10. The y-intercept is at (0, 10)

If you look at finding the y-intercepts in the two problems above, you may see there is a pattern forming. The y-intercept is the number that your adding or subtracting that is located after the x (ex. y = 4x - 2 - the y-intercept would be -2)

9. First, find the x and y-intercepts. The y intercept is -3 You’d graph that at (0, -3). 0 = -1/2x - 3. -1/2x = 3. x = -6 so the x-intercept is at (-6, 0). If you only need to graph 2 points, then you can graph just those two points and draw a line between them.

To graph y=-1/2x + 3, start at the y-intercept and use the slope (-1/2) to find other points. Because your slope is -1/2, you’d go down 1 unit and then to the right 2 units. That would be your next point. If you wanted your line to go further up, go up one unit and then to the left 2 units. That would be your next point.

I am not sure what you need to do on 11 and 12

I think you should try 7, 8 and 10 on your own and let me know if you have any questions on them or if you are stuck on anything.
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