Assume that f(3)=4. Name a point that must be on the graph of y=f(x)+4
(3,8)
if x = 3
then f(x) = 4
so y = 4 + 4 = 8
Since $f(-3)=6$, let $x = -7$. Then $f(-7+4)=6$, and $\boxed{(-7,6)}$ is on the graph.
Alternatively, the graph of $y=f(x+4)$ is a horizontal shift of the graph of $y=f(x)$ by 4 units to the left. Since $(-3,6)$ is on the graph of $y=f(x)$, we know that $\boxed{(-7,6)}$ is on the graph of $y=f(x+4)$.
Well, if we know that f(3) = 4, then we can say that the point (3, 8) must be on the graph of y = f(x) + 4. Why? Because if you substitute x = 3 into the equation, you get y = f(3) + 4 = 4 + 4 = 8. So, the point (3, 8) is a guaranteed member of the graph. Can't promise you any other points though, they may be late to the party!
To find a point on the graph of y = f(x) + 4, we need to know the function f(x) itself. Since we are given that f(3) = 4, it means that when x = 3, the value of f(x) is 4.
Therefore, a point that must be on the graph of y = f(x) + 4 would be (3, 8) since f(3) = 4 and adding 4 to it gives us the y-coordinate of the point on the graph.
In general, to find a point on the graph of y = f(x) + k, where k is a constant, you can evaluate f(x) for a given value of x and then add the constant k to it to find the corresponding y-coordinate.