In both cases there are more than one possible function sutisfying given data.
1. If
- x‑intercepts are (–5, 0), (2, 0), and (6, 0);
- the domain is –5 ≤ x ≤ 7;
- the range is –4 ≤ y ≤ 10,
then (see attached diagram for details) you can build infinetely many functions. From the diagram you can see two graphs: first - blue graph, second - red graph. Translating their maximum and minimum left and right you can obtain another function that satisfies the conditions above.
2. If
- x‑intercepts are (–4, 0) and (2, 0);
- the domain is all real numbers;
- the range is y ≥ –8,
then you can also build infinetely many functions. From the diagram you can see two graphs: first - blue graph, second - red graph. Translating their minimum left and right you can obtain another function that satisfies the conditions above.
Note, that these examples are not unique, you can draw a lot of different graphs of the functions.
Answer: yes, there are more than one possible function
Answer:s^2-25
Step-by-step explanation:
Answer:
3z+35
Step-by-step explanation:
Answer:
w=10
Step-by-step explanation:
8.25+1/4w=10.75
8.25+0.25w=10.75
0.25w=10.75-8.25
0.25w=2.5
w=10
I hope this helps
Solution:
The Preimage function is , F(x)=|x|
Consider a point (a,b) on the function, F(x)=|x|. it means
b=|a|-----(1) satisfies the function.
Now the function F(x)=|x| is vertically stretched by 5 units.It means
x=a , y= b+5
x=a, b=y-5
So, putting the value of a and b in equation (1).
y-5=|x|
y= |x| + 5 is vertical stretch of y= |x| by 5 units.