# Download Calculus with analytic geometry by Harley Flanders; Justin J Price PDF

By Harley Flanders; Justin J Price

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**Extra resources for Calculus with analytic geometry**

**Example text**

By the distance formula, this condi tion is (y + (y - 2)2 4y + 4 = IOy + 25, I +9 - 4y + 5 IOy + 34, y = - 29). Answer The line y = 29). • = = (x,(x, d2 (x,di = d2 , = = (x - 3)2 + 5)2, (x - 1)2 + x2 -6x + y2 + x2 -2x + + y2 -2x = -6x + n(4x n(4x - Midpoints Jn Example I , we found the perpendicular bisector of a line segment PQ by solving a locus problem. There is a different approach: If we knew the midpoint M of PQ, we could write the equation of the line through M and perpendi cular to PQ.

X2• 50 1 . FUNCT I O N S A N D G RAPHS Factor Theorem off (x). f(x) = (x - r)g(x), where g(x) is a polynomial of degree n - 1. 11, and let x = r be a zero I We can interpret the Factor Theorem as a statement about dividing out zeros: If /(x) has degree n and/(r) = 0, then/(x) is divisible by (x - r); hence (x) g(x) = J_" x-r is 2 polynomial of degree n - I. It may happen that the quotient g(x) also has x = r as a zero. In that case x - r can divided out of g(x), so g(x) f (x) x - r (;· _ r)2 is a polynomial.

EXAMPLE 4 Graph y = (x 2)(x l) Sohltion The graph is undefined at x = - 2 and x = 1 . - 2,Thebutfuncti not atonxis=positive 0: • + _ = y•O 1 undefined y>O y