Answer:
known facts:
The graph of y=f(x−a), a>0
y=fx-a, a>0
can be obtained by shifting graph y=f(x)
y=fx, a units right side of x-axis. The graph of y=f(x)+a, a>0 y= fx+ a,a>0
can be obtained by shifting graph y=f(x)
y=fx, a units upwards along y-axis.
Adrienne said:
a vertical stretch by a factor of 4, a horizontal shift to the left 1 unit and a vertical shift down 3 units of the graph of the parent function f(x)=ex
f(x)=e^x
produce the graph of the function g(x)=4ex−1+3
g(x)=4e^x-1+3.
by the above known facts, it can be concluded that
Adrienne is not correct
Step-by-step explanation:
i hope this helps you, i'm still doing it but i have to go for a min, one sec! <3
Adrienne is close, but she mixed up left and right
The x-1 in the exponent means we shift 1 unit right, and not left.
It's a bit confusing since it seems backwards.
But you can think of it like this: replacing x with x-1 shifts the xy axis 1 unit to the left, so we have the illusion the curve moved 1 unit to the right.
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Here's one way to write out the transformation steps
f(x) = e^x
4*f(x) = 4*e^x .... vertical stretch by factor of 4
4*f(x-1) = 4*e^(x-1) ... shift 1 unit right
4*f(x-1)+3 = 4*e^(x-1)+3 ... shift 3 units up
g(x) = 4*e^(x-1)+3
