Translations: y = x^2 to y = a(x - p)^2 + q
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mmm4444bot
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The rules for horizontal shifts are opposite of what we would expect.
A graph of function f(x) shifts to the right, when we subtract a constant from the input x.
f(x-4) shifts the graph four units to the right.
The graph shifts to the left, when we add a constant to the input x.
f(x+4) shifts the graph of f(x) four units to the left.
One way to think about this is in terms of shifting the coordinate axes, instead of the graph. In other words, x+4 shifts the axes 4 units to the right, so the graph gets left behind. If you slide the coordinate system to the right without sliding the graph also, the end effect is the same as leaving the axes in place and instead shifting the graph to the left by the same amount.
Try it, with a transparency on top of xy-axes. Draw the graph on the transparency. Then hold the transparency still, while sliding the axes underneath right four units. The graph ends up shifted to the left.
You can also google keywords like why does adding shift graph left, to find more explanations.
A graph of function f(x) shifts to the right, when we subtract a constant from the input x.
f(x-4) shifts the graph four units to the right.
The graph shifts to the left, when we add a constant to the input x.
f(x+4) shifts the graph of f(x) four units to the left.
One way to think about this is in terms of shifting the coordinate axes, instead of the graph. In other words, x+4 shifts the axes 4 units to the right, so the graph gets left behind. If you slide the coordinate system to the right without sliding the graph also, the end effect is the same as leaving the axes in place and instead shifting the graph to the left by the same amount.
Try it, with a transparency on top of xy-axes. Draw the graph on the transparency. Then hold the transparency still, while sliding the axes underneath right four units. The graph ends up shifted to the left.
You can also google keywords like why does adding shift graph left, to find more explanations.
You said "f(x+4) shifts the graph of f(x) four units to the left." So in this case the question asks for a translation four units left. Where I'm confused is I chose the answer p=4 and that was incorrect. The correct answer is supposedly p=-4; but wouldn't that be a translation of four units right?
Dr.Peterson
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But the equation is y = a(x - p)^2 + q; p is not being added, but subtracted! So p is the amount of shift to the right, not to the left; a negative p shifts to the left.
When you put p = -4 into the equation, you get y = 2(x - -4)^2 + 3 = 2(x + 4)^2 + 3, which is what you expect.
It's important to read carefully. The form they are using is common; the new vertex is (p,q), and the subtraction makes this possible.
Don't...best way to learn! Now, it'll never happen again
mmm4444bot
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You're in good company.
One of the most common mistakes students make is to think that notation like -c represents a negative number (everytime they see it). After all, there's a negation sign in front of symbol c, so -c must be negative. Right?
Not right, heh. We need to be mindful of different cases, when working with symbolic expressions.
The expression -c could represent a positive number, a negative number, or neither. Without context, we just don't know.
As Jeff explained, the subtraction x - c tells us that the opposite of value c is being added.
x - c = x + (-c)
This means that x is either getting larger, or smaller, or not changing at all. Again, we can't say which, until after we know the value of c.
x - c
Case 1: x gets smaller, when c > 0
Case 2: x gets larger, when c < 0
Case 3: x does not change, when c = 0
Your issue concerned case 2.