^{2}and y= (x-4)

^{2 }.

^{2}since y = 1(x-0)

^{2}+ 0.

**right**of 4.

^{2}.

^{2}+ 0. So the vertex was moved 4 units to the left from (0, 0) to (-4, 0).

Example: In –x^2– 4x+ 5 I would say ok so it has a reflection of the y axis, It moves right 4 units and moves up 5. But the answer is that instead of moving 4 units right, its 4 units LEFT. Why???

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Take y = x^{2} and y= (x-4)^{2 }.

The parabolic form is y = a(x MINUS h) PLUS k, where (h, k) is the vertex of the parabola opening up or down depending on the sign of a.

So the vertex is at (0,0) for y = x^{2} since y = 1(x-0)^{2} + 0.

Now, for y = (x-4)2, note that the vertex (which is essentially the "handle" for the parabola) is now at

(+4, 0).

So despite the (x MINUS 4) being squared, the x coordinate of the vertex was shifted from 0 to 4, which is a shift to the
**right** of 4.

Now consider y=(x+4)^{2}.

That's really y = (x - (-4))^{2} + 0. So the vertex was moved 4 units to the left from (0, 0) to (-4, 0).

Hope this helps.

For quadratic functions such as this one, you should complete the square to see how the parabola is shifted and reflected. In your case,

f(x) = –x²– 4x+ 5 = -(x+2)² + 9

The +9 tells us the parabola is shifted up (positive y-direction) by 9 units relative to the origin, the +2 tells us it is shifted left by 2 units (negative x-direction), so the vertex is at (-2,9). The negative sign tells us it opens down (in the negative y-direction).

More generally, if you change a function from f(x) to f(x+A) for any positive constant A, its graph is shifted A units to the left. f(x-A) shifts the graph A units to the right. If you change f(x) to f(x)+A, it shifts it A units up, and f(x)-A is shifted A units down.

Y=x^2 is centered around origin.

Assuming your coordinates are traditional (+ to the right and - to the left of the origin)

Analyze y=x^2 - 4x:

To bring y back to the origin, we have to add (move 4x units to the right).

This means y moved -4x units which is 4x to the left.

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