1.Describe how the graph of y= x2 can be transformed to the graph of the given equation. y = x2 - 14 Shift the graph of y = x2 right 14 units. Shift the graph of y = x2 up 14 units. Shift the graph of y = x2 left 14 units. Shift the graph of y = x2 down 14 units.
2.Describe how the graph of y= x2 can be transformed to the graph of the given equation. y = (x-14)2 - 9 Shift the graph of y = x2 down 14 units and then left 9 units. Shift the graph of y = x2 right 14 units and then up 9 units. Shift the graph of y = x2 right 14 units and then down 9 units. Shift the graph of y = x2 left 14 units and then down 9 units. 3.Describe how to transform the graph of f into the graph of g. f(x) = x4 and g(x) = -x4 Reflect the graph of f across the x-axis. Shift the graph of f down 1 unit. Reflect the graph of f across the x-axis and then reflect across the y-axis. Reflect the graph of f across the y-axis. 4.The transformation from f to g represents a __________ stretch. f(x) = and g(x) = 6 Note: Use all lowercase letters in your response. 5.Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x. f(x) = x3 + 4 and g(x) =
In general, if you have y = x^2 and you add some number to it (say 5), then you are moving the graph up (5 units) If you subtract a number from it (again say 5), you are shifting the graph down (5 units)
okay, thank you
what about for the last three?
you're welcome
I'll do 2 more, but you should really post each question individually (it will save space and draw in more people to help)
y = x^2 is your parent graph, which is your standard parabola (opening upward) if you replace x with x-14, you are effectively moving the graph 14 units to the right (not left). Notice how when x = 0, then y = x^2 turns into y = 0. Then when x = 14, y = (x-14)^2 turns into y = 0. So that proves we have moved 14 units to the right
Finally, we're subtracting 9 from the whole thing. So we shift the whole graph down 9 units
making sense so far?
ish?
where are you stuck
okay i got it now, i had to reread what your saying.. i need help with 3 and 5, i understand the other ones now, (thanks to you)
If you have any function, and you stick a negative sign outside the entire function, then you are changing the y coordinates of the graph from negative to positive (or vice versa)
So say you start with y = x^2 The y coordinates are all positive on this graph but if you stick a negative sign out front, you get y = -x^2 and that turns all the positive y coordinates negative
So what happens is that you get a reflection over the x axis
for instance, say you had these points that make up this parabola |dw:1364499575464:dw|
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