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the graph of the function g(x) has the same shape and direction of opening as the graph of f(x)=3(x-2)^2+9. the graph ofg(x) has a vertex that is 2 units to the right and 5 units down from the vertex of the graph of f(x)

 

3 years ago

Answered By Emily D

Since I'm not sure where you got stuck on this question, I'm going to walk you through it like this topic is brand new: they gave us f(x) in vertex form. Vertex form always looks like f(x) = A(x-p)2 + q, where A, p, and q are regular old numbers

A - tells us how much the graph is stretched vertically and which direction it opens in (+ up, - down)

p - tells us how much the graph is shifted left or right (- right, + left), notice the positive/negative are BACKWARDS from what we would normally expect

q - tells us how much the graph is shifted up or down (+ up, - down)

If you look at (-p,q) as coordinates, that is where the vertex (the max or min) sits on your graph

 

Let's look at the function we were given: f(x) = 3(x-2)2 + 9 and compare it to y = x2

A = 3, the graph is stretched 3 times in the up-down direction

p = 2, the graph's vertex is 2 units to the RIGHT of where x2 would be

q = +9, the graph's vertex is 9 units UP from where x2 would be

f(x) has a vertex at (2,9)

 

so now let's investigate what we know about g(x)

- has the same shape and direction opening: this tells us that A is the same for both

- vertex is 2 units to the right: p is 2 more than it was in f(x), so it goes from x = 2 to x = 4

- vertex is 5 units down: q is 5 less than it was in f(x), so it goes from y = 9 to y = 4

g(x) = 3(x - 4)2 + 4