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The function will simply get reflected about the y-axis.
Let's approach this through what we know. Since we know that the x values are mirrored, we know that the points in Quadrant I and IV will be reflected over to the negative side, Quadrants II and III, because they simply change in signs.
However, we also know that the function y-values do not change. This is because whatever the x values are don't change the range and y-values of an even function.
To be more specific, if we have an even function, we are most likely dealing with quadratics or variants/transformations of the quadratic function.
If we were to have 2, and -2, and we wanted to plug them into the equation:
, the signs do not change the y-values of the function.
Hence, we know that it ONLY gets reflected across the y-axis.
Let's approach this through what we know. Since we know that the x values are mirrored, we know that the points in Quadrant I and IV will be reflected over to the negative side, Quadrants II and III, because they simply change in signs.
However, we also know that the function y-values do not change. This is because whatever the x values are don't change the range and y-values of an even function.
To be more specific, if we have an even function, we are most likely dealing with quadratics or variants/transformations of the quadratic function.
If we were to have 2, and -2, and we wanted to plug them into the equation:
Hence, we know that it ONLY gets reflected across the y-axis.
Answer:
Step-by-step explanation:
The following equation y=f(x+3) can be seen graphed in the image attached to this answer. The point P(-6,-2) shows up if we input -6 as the variable x which will output -2 as the value for y. This can be seen in the graph below.
y=f(x+3)
y= -6 + 3
y = -2
