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 March 9th, 2011, 03:46 PM #1 Newbie   Joined: Mar 2011 Posts: 1 Thanks: 0 points of intersections Hey guys I know this is going to seem like a retarted question but here goes. determine the point of intersection of f(x)=5+?x+1 and g(x)=x+1 when you set them equal the equation it ends in is 0=x^2-9x+15 how do i resolve for the x and y values for the point of intersection?
 March 9th, 2011, 03:58 PM #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 Re: points of intersections Are you sure you have f(x) copied correctly? I ask because I get a different equation in x when I equate f and g, and it seems odd it would be given with like terms left to combine. My best guess is that you are given: $f(x)=5+\sqrt{x}$ $g(x)=x+1$ Equate f and g: $5+\sqrt{x}=x+1$ Subtract 5 from both sides: $\sqrt{x}=x-4$ Square both sides, keeping in mind extraneous solutions may be introduced, if they are less than 4 (a square root cannot be negative): $x=x^2-8x+16$ Write in standard quadratic form: $x^2-9x+16=0$ Apply quadratic formula: $x=\frac{9\pm\sqrt{9^2-4\cdot16}}{2}=\frac{9\pm\sqrt{17}}{2}$ Discard root less than 4: $x=\frac{9+\sqrt{17}}{2}$ Now, for the y-coordinate use either $f$$\frac{9+\sqrt{17}}{2}$$$ or $g$$\frac{9+\sqrt{17}}{2}$$$.
March 10th, 2011, 08:32 AM   #3
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Re: points of intersections

Hello, functionman!

Quote:
 $\text{Determine the point of intersection of: }\: f(x)\:=\:5\,+\,\sqrt{x+1}\,\text{ and }\,g(x)\:=\\,+\,1" /> $\text{When you set them equal, the equation becomes: }\^2\,-\,9x\,+\,15 \:=\:0" /> $\text{How do I solve for the }x\text{ and }y\text{ values for the point of intersection?$

$\text{Quadratic Formula: }\;x \;=\;\frac{-(-9)\,\pm\,\sqrt{(-9)^2\,-\,4(1)(15)}}{2(1)} \;=\;\frac{9\,\pm\,\sqrt{21}}{2}$

$\text{Substitute into }g(x):\;\;y \;=\;\frac{9\,\pm\,\sqrt{21}}{2}\,+\,1 \;=\;\frac{11\,\pm\,\sqrt{21}}{2}$

$\text{The points are: }\;\left(\frac{9+\sqrt{21}}{2},\;\frac{11+\sqrt{21 }}{2}\right)\,\text{ and }\,\left(\frac{9-\sqrt{21}}{2},\;\frac{11-\sqrt{21}}{2}\right)$

 March 10th, 2011, 08:46 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: points of intersections I guess your years as a professor has better equipped you to translate what people mean to write vs. what they actually write!

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