# Determine x- and y-intercepts of the graph write as points also the domain and vertical asymptotes if any of function write domain in interval notation. f(x)=x+4/x2-25Determine the horizontal or...

Determine x- and y-intercepts of the graph write as points also the domain and vertical asymptotes if any of function write domain in interval notation. f(x)=x+4/x2-25

Determine the horizontal or oblique asymptote if any of function obtain additional points on the graph use a table to show work in finding the additional points plot the x- and y- intercepts, the additional points, and the asymptotes that you found on a rectangular coordinate system graph the function.  Draw the asymptotes using dashed lines.  Submit graph f(x)=x+4/x2-25.

nathanshields | High School Teacher | (Level 1) Associate Educator

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y-intercept: let x=0.  Then y=-4/25

x-intercepts: let y=0. Then x=-4

End behavior: since degree of denominator is larger, the function tends toward 0 at the extreme x values.  Horizontal asymptote y=0.

Vertical asymptotes (or holes) occur when we divide by 0.  `x^2-25=0->x=+-5`

So the domain is all numbers except -5 and 5.  Interval notation:

`(-infty,-5)uu(-5,5)uu(5,infty)`

To find some points on the graph, just choose random x-values.

(1, -5/24)

(6, 10/11)

etc.

rcmath | High School Teacher | (Level 1) Associate Educator

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X-intercept: Let y=0, then (x+4)/(x^2-25)=0 ==> x+4=0 ==>x=-4    (-4,0)

y-intercept: Let x=0, then y=4/-25 so (0,4/-25)

This function is defined if the denominator is not zero so x^2-25not zero which means x is not 5 or -5                                                   so domain=(-inf,-5)U(-5,5)U(5,inf)

Since 5 and -5 makes the denominator zero then your vertical asymptotes are x=-5 and x=5.