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August 17th, 2014, 05:43 AM   #1
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Attached is the question

i absolutely had so much trouble with this one.

there is no answer in the back of the book for part 7 a.

however, the answer for 7 b is:

3 1/8 units (this is a fraction, 3 and 1/8 units)
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August 17th, 2014, 08:22 AM   #2
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For part a) of the question, we need to equate the two functions and then solve for $x$:

$\displaystyle x^2-3x=2x-x^2$

Move everything to the left:

$\displaystyle 2x^2-5x=0$

Factor:

$\displaystyle x(2x-5)=0$

Now, using the zero-factor property, we equate each factor in turn to zero and solve for $x$:

$\displaystyle x=0$

$\displaystyle 2x-5=0\implies x=\frac{5}{2}$

Thus, we have shown that the two graphs intersect for:

$\displaystyle x\in\left\{0,\frac{5}{2}\right\}$

For part b), we may describe the vertical separation between the two graphs on the given interval as the difference between the upper graph and the lower graph:

$\displaystyle d(x)=\left(2x-x^2\right)-\left(x^2-3x\right)=5x-2x^2=x(5-2x)$

Now, this resulting quadratic function has two roots:

$\displaystyle x\in\left\{0,\frac{5}{2}\right\}$

Because we see the squared term has a negative coefficient, we know the vertex will be its maximum. And we know the axis of symmetry will lie midway between the two roots. Thus the axis of symmetry is:

$\displaystyle x=\frac{5}{4}$

And so the maximum vertical separation is:

$\displaystyle d_{\max}= d\left(\frac{5}{4}\right)= \frac{5}{4}\left(5-\frac{5}{2}\right)= \frac{5}{4}\cdot\frac{5}{2}= \frac{25}{8}$
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Last edited by skipjack; August 17th, 2014 at 10:24 AM.
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August 17th, 2014, 10:42 AM   #3
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Let f(x) = x² - 3x and g(x) = 2x - x², so the graphs are of y = f(x) and y = g(x).

As f(0) = g(0) = 0 and f(2$\small\frac12$) = g(2$\small\frac12$) = -1$\small\frac14$, the graphs meet where x = 0 and x = 2$\small\frac12$.

The vertical separation is |f(x) - g(x)| = |2x² - 5x|, and the previous post shows how to proceed from there.
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