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.

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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.**

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.

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