Me and if you explain good and i get it right i will give a

Me and if you explain good and i get it right i will give a


[tex]Me and if you explain good and i get it right i will give a[/tex]

Related Posts

This Post Has 4 Comments

  1. The answer is A because all of the coordinates match to the points up above. Lets take the coordinate (-3,6) for example. -3 would be where the point is on the x-axis which is horizontal (side to side), that means the 6 would be on the y-axis¬†vertical(up and down). Hope this helps ūüôā

  2. Hey there, and good morning to you!

    So, I want to go a bit in depth on how to find points so you can figure it out on your own next time!

    We have the x-axis and the y-axis. The x-axis is horizontal, and the y-axis is vertical. Point is put into this quotation: (x,y)

    Where the point lies on the horizontal line is x, and where it lies on the vertical line is y. Keep in mind that some quadrants have negative or positive coordinates. Keep track of the number line!

    Let's look at the highest point. 

    If I drew a straight line all the way down, I would be on -3 on the x-axis. That is our x. If I drew a line across from that point, I would be on point 6, that is our y. 

    (-3,6) <== first point. 

    Try the next one. 

    x is on point 3,  we know the x is point 3. There is no y since it is lying on the x-axis. y= 0

    (3,0) <== second point

    Last one:

    (-1, -6) If I drew I line up, I would be at -1. If I drew a straight line across, I would be at -6. 

    "D" is your correct answer.

    I hope this helps!
    ~kaikers

  3. g(x) = x² + 2

    step-by-step explanation:

    it is given that the graphs are the same shape.

    the only difference is that the red graph is 2 units higher than the blue graph.

    that is to say that the red graph is simply the blue graph that has been moved vertically by 2 units in the positive y direction.

    hence

    g(x) = f(x) + 2

    or

    g(x) = x² + 2

    [tex]The graphs below have the same shape. what is the equation of the red graph?[/tex]

Leave a Reply

Your email address will not be published. Required fields are marked *