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February 18th, 2019, 08:48 PM   #1
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Graph of a function


The graph of the function f(x) = a(x– 2)^2(x+ b) is shown in the figure.Find a and b.
I solved this question by plugging f(2)=0 then I get b=-2
Then I plugged f(0)=3 and i found a but answer is not true according to my book why is this so??
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February 18th, 2019, 09:10 PM   #2
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$f(2)=0$ regardless of what the value of $b$ is. Do you see why?

How about if we note that

$f(0)=3$

$4ab = 3$

They then seem to be trying to indicate that the maximum of $f(x)$
occurs at $x=3$

If this is the case then the first derivative of $f(x)$ will be equal to $0$ at $x=3$

$\dfrac{df}{dx} = 2a(x-2)(x+b) + a(x-2)^2 = (x-2)(2a(x+b)+a(x-2))$

$2a(x+b)+a(x-2) = 0 ~@x=3$

$2a(3+b) + a(3-2) = 0$

$6a+2ab+a = 0$

$a(7+2b)=0$

$a=0$ or $b = -\dfrac{7}{2}$

$a \neq 0$ as $4ab=3$ from above

so $b = -\dfrac{7}{2}$

$4 a \left(-\dfrac{7}{2}\right) = 3$

$-14a = 3$

$a = -\dfrac{3}{14}$
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February 18th, 2019, 10:13 PM   #3
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Quote:
Originally Posted by romsek View Post

They then seem to be trying to indicate that the maximum of $f(x)$
occurs at $x=3$
This should read they seem to be trying to indicate that there is a local maximum of $f(x)$ at $x=3$
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