finding domain + values of a,b for a rational function

[fresh]

y=[(2x^2)-8]/[(ax^2)+(bx)-10]


This function has only one vertical asymptote, whose equation is x= 5 . It is also known that the function is undefined for one more x value, which is negative

a) domain
for the domain, i assume i make ax^2)+(bx)-10 = 0 and solve. but how do i solve this without knowing values for a and b? is there another way to find domain?

b)a,b
since there is a vertical asymptote when x=5
(ax^2)+(bx)-10 =0 when x =5
25a+5b-10 = 0
5a+b-2=0
b=2-5a

where do i go from here?

 

[Dr.Peterson]

y=[(2x^2)-8]/[(ax^2)+(bx)-10]

This function has only one vertical asymptote, whose equation is x= 5 . It is also known that the function is undefined for one more x value, which is negative

a) domain
for the domain, i assume i make ax^2)+(bx)-10 = 0 and solve. but how do i solve this without knowing values for a and b? is there another way to find domain?

b)a,b
since there is a vertical asymptote when x=5
(ax^2)+(bx)-10 =0 when x =5
25a+5b-10 = 0
5a+b-2=0
b=2-5a

where do i go from here?

When I think about rational functions, I focus my attention on factors. What factor do you know the denominator must have? You also know that some (other) factor must be in both the numerator and the denominator.

Then factor the numerator fully, and apply what you know.

 

[fresh]

When I think about rational functions, I focus my attention on factors. What factor do you know the denominator must have? You also know that some (other) factor must be in both the numerator and the denominator.

Then factor the numerator fully, and apply what you know.

factoring the numerator i get 2x^2 - 8= 2(x^2-4)=2(x+2)(x-2)
in order for there to be an undefined value at a negative number, the denominator would need the factor (x+2), so that there would be a hole at x+2

so the function would be 2(x+2)(x-2)/(x-5)(x+2)
and the domain would be all real numbers except x cannot equal x=-2, 5 ?

then to find a,b
(ax^2)+(bx)-10 =0 when x =5
25a+5b-10 = 0
5a+b-2=0
b=2-5a

and
(ax^2)+(bx)-10 =0 when x =-2
a(-2)^2+b(-2)-10=0
4a-2b-10=0
2a-b-5=0
b=2a-5

equating the two,
2-5a=2a-5
2+5=2a+5a
7=7a
a=1

b=2(1)-5
=-3

so that b=-3 and a=1?

 

[Dr.Peterson]

factoring the numerator i get 2x^2 - 8= 2(x^2-4)=2(x+2)(x-2)
in order for there to be an undefined value at a negative number, the denominator would need the factor (x+2), so that there would be a hole at x+2

so the function would be 2(x+2)(x-2)/(x-5)(x+2)
and the domain would be all real numbers except x cannot equal x=-2, 5 ?

then to find a,b
(ax^2)+(bx)-10 =0 when x =5
25a+5b-10 = 0
5a+b-2=0
b=2-5a

and
(ax^2)+(bx)-10 =0 when x =-2
a(-2)^2+b(-2)-10=0
4a-2b-10=0
2a-b-5=0
b=2a-5

equating the two,
2-5a=2a-5
2+5=2a+5a
7=7a
a=1

b=2(1)-5
=-3

so that b=-3 and a=1?

Good work, but there's a much easier way to find a and b.

You know that the denominator is (x-5)(x+2), or perhaps some multiple of that. Expanding it, we get x^2 - 3x - 10. If this is to equal ax^2 + bx - 10, then a must be 1, and b must be -3. Just match up the coefficients!

Note that if they had said it was ax^2 + bx + 20, then it would have to be -2 times what you thought, in order to get the right constant term, namely -2(x-5)(x+2) = -2(x^2 - 3x - 10) = -2x^2 + 6x + 20, giving a = -2 and b = 6. So your approach would have needed a little tweaking. To account for this possibility, it would have been wiser to say that the function must be [2(x+2)(x-2)]/[a(x-5)(x+2)], not necessarily [2(x+2)(x-2)]/[(x-5)(x+2)].