Question

How do you graph `y=(3x^2+10x-8)/(x^2+4)` using asymptotes, intercepts, end behavior?

Answer 1

See calculations below

Explanation:

The domain of y is `D_y =RR`

The denominator is `x^2+4>0`, `AAx in RR`

To find the intercepts, let `x=0`

`y=-8/4=-2`

The y-intercept is `(0,-2)`

When `y=0`

`3x^2+10x-8=0`

`(3x-2)(x+4)=0`

So, `x=2/3` and `x=-4`

The x-intercepts are `(2/3,0)` and `(-4,0)`

To calculate the limits when `x->+-oo`, we take the terms of highest degree in the numerator and the denominator.

`lim_(x->+-oo)y=lim_(x->+-oo)(3x^2)/x^2=3`

So, the horizontal asymptote is `y=3`

When `y=3`, `=>`, `3=(3x^2+10x-8)/(x^2+4)`

`3x^2+12=3x^2+10x-8`

`10x=20`, `=>`. `x=2`

The curve cuts the horizontal asymptote at `(2,3)`

To find the maximum and minimum, you have to calculate the derivative.

`y=u/v`

`dy/dx=(u'v-uv')/v^2`

`u=3x^2+10x-8`, `=>`. `u'=6x+10`

`v=x^2+4`, `=>`, `v'=2x`

`dy/dx=((6x+10)(x^2+4)-(2x)(3x^2+10x-8))/(x^2+4)^2`

`=(6x^3+24x+10x^2+40-6x^3-20x^2+16x)/(x^2+4)^2`

`=(-10x^2+40x+40)/(x^2+4)^2`

`dy/dx=0`

`-10(x^2-4x-4)=0`

`Delta=b^2-4ac=16+16=32`

`x=(4+-sqrt32)/2=(4+-4sqrt2)/2=2+-2sqrt2`

So we have 2 points, `x=2+2sqrt2` and `x=2-2sqrt2`

We do a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `2-2sqrt2` `color(white)(aaaa)` `2+2sqrt2` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `dy/dx` `color(white)(aaaaaaaa)` `-` `color(white)(aaaaaaaa)` `+` `color(white)(aaaaaaa)` `-`

`color(white)(aaaa)` `y` `color(white)(aaaaaaaaaa)` `darr` `color(white)(aaaaaaaaa)` `uarr` `color(white)(aaaaaaa)` `darr`

`x=2-2sqrt2` corresponds to a minimum

`x=2+2sqrt2` corresponds to a maximum

graph{(y-(3x^2+10x-8)/(x^2+4))(y-3)=0 [-10, 10, -5, 5]}

Answer 2

x-intercepts `(y=0): -4 and 2/3`; y-intercept `(x=0): -2`. Horizontal asymptote: `larr y = 3 rarr. y in (-3.04, 4.04)`, nearly.

Explanation:

x-intercepts `(y=0): -4 and 2/3`; y-intercept `(x=0): -2`

By actual division,

`y =3+(10(x-2))/(x^2+4) to (2, 3)` is a point on the graph.

`y=3+(10(2-1/x))/(x(1+4/x^2)) to 3`, as x to +-oo#.

It is evident that y = 3 is the horizontal asymptote that cuts the graph

at (2, 3)..

`y'=(-10x^2+40x+40)/(x^2+1)^2= 0`, when x = 2(1+-sqrt2)=-.83 and

4.83,

nearly. These turning points are (-.93, -3.04) and (4.83, 4.04), nearly.

The graph justifies mini/max y at these points as -3.04/4.04,

nearly. y is bounded between these extremes.

graph{y(x^2+4)-3x^2-10x+8=0 [-16.29, 16.24, -8.13, 8.14]}