# Chapter 4 - Section 4.3 - How Derivatives Affect the Shape of a Graph - 4.3 Exercises - Page 303: 76

$a=-\frac{20}{3}$ $b = \frac{4}{3}$ The other inflection points are $(0,0)$ and $(-2, -2.5)$

#### Work Step by Step

$x^2y+ax+by=0$ $y(x^2+b) = -ax$ $y = -\frac{ax}{x^2+b}$ We can find $y'$: $y' = -\frac{a(x^2+b)-(ax)(2x)}{(x^2+b)^2}$ $y' = -\frac{ax^2+ab-2ax^2}{(x^2+b)^2}$ $y' = \frac{ax^2-ab}{(x^2+b)^2}$ We can find $y''$: $y'' = \frac{(2ax)(x^2+b)^2-(ax^2-ab)(2)(x^2+b)(2x)}{(x^2+b)^4}$ $y'' = \frac{(2ax)(x^2+b)-(ax^2-ab)(2)(2x)}{(x^2+b)^3}$ $y'' = \frac{2ax^3+2abx-4ax^3+4abx}{(x^2+b)^3}$ $y'' = \frac{6abx-2ax^3}{(x^2+b)^3}$ To find the points of inflection, we can find values of $x$ such that $y'' = 0$: $y'' = \frac{6abx-2ax^3}{(x^2+b)^3} = 0$ $6abx-2ax^3 = 0$ $-2ax(x^2-3b) = 0$ $x = 0$ or $x^2 = 3b$ $x = 0$ or $x = \pm \sqrt{3b}$ It is given that $(2,2.5)$ is an inflection point. $x = \sqrt{3b} = 2$ $3b = 4$ $b = \frac{4}{3}$ We can find $a$: $x^2y+ax+by=0$ $(2)^2(2.5)+2a+(\frac{4}{3})(2.5)=0$ $10+2a+\frac{10}{3}=0$ $2a=-10-\frac{10}{3}$ $2a=-\frac{40}{3}$ $a=-\frac{20}{3}$ We can find the other inflection points: When $x = 0$: $x^2y+ax+by=0$ $(0^2)y+a(0)+\frac{4}{3}y=0$ $0+0+\frac{4}{3}y=0$ $y = 0$ When $x = -2$: $x^2y+ax+by=0$ $(-2)^2y+(-\frac{20}{3})(-2)+\frac{4}{3}y=0$ $4y+\frac{40}{3}+\frac{4}{3}y=0$ $12y+40+4y=0$ $16y = -40$ $y = -\frac{40}{16}$ $y = -2.5$ The other inflection points are $(0,0)$ and $(-2, -2.5)$

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