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$y=f(x)$ Maximum point: $(2,-16)$

Then asked to find maximum values for different functions;

$y=f(x)+6$

$y=f(x-2)$

$y=f(4x)$

I'm not sure how to determine the maximum or minimum value of a function when it is not a quadratic.

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    $\begingroup$ How do you maximise $f(x)+6$? What is a simpler expression to maximise? $\endgroup$ – Element118 Nov 7 '15 at 12:59
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These are just transformations of functions.

$f(x) + 6$ is just the function shifted upwards (or shifted parallel to the $y$-axis in the positive direction) by $6$ units, so the new maximum is at $(2, -10)$

$f(x-2)$ is a horizontal shift (or shift parallel to the $x$-axis in the positive direction) by 2 units so the new maximum is at $(4, -16)$

$f(4x)$ is a stretch parallel to the $x$-axis, scale factor $\frac{1}{4}$ so the new maximum is at $(\frac{1}{2}, -16)$

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  • $\begingroup$ The maximum of $f$ is given to be $(2,-16)$ and not $(-2,-16)$ $\endgroup$ – marwalix Nov 7 '15 at 13:08
  • $\begingroup$ Yes, I've corrected it. $\endgroup$ – Zain Patel Nov 7 '15 at 13:09

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