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If I wanted to change the vertex of a quadratic function from (2,3) to (-1,6), what form would you use and what would you do to it? (Be specific - Consider the form of the equation)

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Ideally you would want a function in vertex form:
f(x)=a(x-h)^2+k with (h,k) as the vertex coordinates

with a vertex at (2,3) it would look like this:
f(x)=a(x-2)^2+3
to change it to (-1,6) change it to:
f(x)=a(x+1)^2+6

if you have a quadratic function in general form it looks like this:
f(x)=y=ax^2+bx+c
then you would have to transform it first with the given a,b,c values to the vertex form or you just apply general transformations/shifts of the function

for general shifts/transformations you can modify the y coordinate by subtracting or adding a constant amount to the function
so again if we have the vertex (2,3) and want to move it to (-1,6)
change y: f(x)=ax^2+bx+c
new vertex is 3 higher than older-> f(x)=g(x)+3
g(x)=ax^2+bx+c+3

change x:
to shift it horizontally you would replace all x, in the example you want to move the function 3 to the left/towards -x

if x is your current position, then x+1 is the next/increased/further towards +x direction. If you insert this value instead into your function then this value one to the right f(x+1) becomes your current f(x) value.
So if you replace all occurrences of x with (x+1) you would shift the function by 1 to the left/-x.

So for a shift of 3 towards -x you would replace all x with (x+3), sort of calling f(x+3) instead of f(x):
f(x+3)=g(x)=a(x+3)^2+b(x+3)+c

both shifts (x-=3,y+=3) in a single transformation:
f(x+3)+3=g(x)=a(x+3)^2+b(x+3)+c+3

The advantage of using transformations like these it that these work for any function or form. It doesn't matter if they are linear, quadratic or something else.
User Bendewey
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