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 Algebra Pre-Algebra and Basic Algebra Math Forum

August 17th, 2014, 05:43 AM   #1
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graphs

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)
Attached Images Untitled picture.jpg (17.9 KB, 7 views) August 17th, 2014, 08:22 AM #2 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 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: $\displaystyle d_{\max}= d\left(\frac{5}{4}\right)= \frac{5}{4}\left(5-\frac{5}{2}\right)= \frac{5}{4}\cdot\frac{5}{2}= \frac{25}{8}$ Thanks from jessjans11 Last edited by skipjack; August 17th, 2014 at 10:24 AM. August 17th, 2014, 10:42 AM #3 Global Moderator   Joined: Dec 2006 Posts: 21,036 Thanks: 2274 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. Thanks from jessjans11 Tags graphs Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post rhubarb123 Pre-Calculus 1 May 16th, 2014 08:23 AM ungeheuer Math Software 2 February 4th, 2013 08:34 AM MageKnight Applied Math 0 January 17th, 2013 10:38 PM johnni370 Algebra 3 January 2nd, 2013 12:15 AM ungeheuer Calculus 2 December 31st, 1969 04:00 PM

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