Cfrgtkky

How do I sketch a graph of the rational function: $r(x) = frac{x^2+6x-8x}{2x^2+10x+12}$ and how do I find the x-intercepts, Y intercept, vertical asympyotes, and horizontal asympyotes.

I know the basic definition of the terms, but this is my first time doing a problem like this and I really want to see if anyone could break it down.

The x-intercepts are the points $x$ where $r(x)=0$. To find these points set the equation equal to 0 and solve for $x$:
$$frac{x^2+6x-8x}{2x^2+10x+12}=0$$
This is 0 when the numerator is 0, so solve the quadratic $x^2+6x-8x=0$ to obtain the x-intercepts.

The y-intercept is the point where the function $r$ crosses the y-axis. This is when $x=0$, so evaluate $r(0)$ to determine the y-intercept.

Because the polynomials in the numerator and denominator have the same degree, then the horizontal asymptote is the ratio of the leading coefficients for the numerator and denominator, in this case $1/2$ and the horizontal asymptote is at $y=1/2$.

The vertical asymptotes occur when the denominator obtains the value 0. So set the quadratic in the denominator equal to 0 and solve for $x$, these values of $x$ are the vertical asymptotes.

Why not simply plot it?

The zeros are when $x^2 + 6 x - 8 = 0$ and the divergences are when $2 x^2 + 10 x + 12 = 0$, and the limit is $x^2/(2 x^2) = 1/2$.

First, eliminate any common root of the numerator and denominator (only for finding the vertical asymptotes). Then find the roots of the denominator. They are all candidates for vertical asymptotes. Also find the limit of the function for $xto pm infty$ to find horizontal asymptotes. To find X-intercepts find $x$s for which $f(x)=0$ and similarly, Y-intercept can be found by substituting $x=0$ into the equation of the function. Clarifying all of this we have $$f(x)={x^2+6x-8over 2(x+2)(x+3)}$$hence, $x=-2$ and $x=-3$ are the vertical asymptotes since they are the roots of the denominator (and not that of the numerator). By tending $xto pminfty$ we also have $y={1over 2}$ as our only horizontal asymptote. The only Y-intercept is $$(0,-{2over 3})$$and the X-intercepts are $$x=-3pm sqrt{17}$$here is a sketch of the function

Getting there

$$
begin{array}{c|c|c|c}
x & x^2 + 6x-8 & 2 x^2 + 10 x + 12 & frac{x^2 + 6x-8}{2 x^2 + 10 x + 12} \ hline
-15&127&312&0.407 \
-14&104&264&0.3939 \
-13&83&220&0.3773 \
-12&64&180&0.3555 \
-11&47&144&0.3264 \
-10&32&112&0.2857 \
-9&19&84&0.2262 \
-8&8&60&0.1333 \
-7&-1&40&-0.025 \
-6&-8&24&-0.3333 \
-5&-13&12& -1.0833 \
-4&-16&4& -4 \
-3.75&-16.4375&2.625&-6.2619 \
-3.5&-16.75&1.5&-11.1667 \
-3.25&-16.9375&0.625&-27.1 \
-3&-17&0& mbox{vert asymp} \
-2.875&-16.984375&-0.21875&77.6428 \
-2.75&-16.9375&-0.375&45.1667 \
-2.625&-16.859375&-0.46875&35.9667 \
-2.5&-16.75&-0.5&33.5 \
-2.375&-16.609375&-0.46875&35.4333 \
-2.25&-16.4375&-0.375&43.8333 \
-2.125&-16.2343&-0.21875&74.2142 \
-2&-16&0& mbox{vert asymp} \
-1.75&-15.4375&0.625&-24.7 \
-1.5&-14.75&1.5&-9.8333 \
-1.25&-13.9375&2.625&-5.3095 \
-1&-13&4&-3.25 \
0&-8&12&-0.6667 \
1&-1&24&-0.0417 \
2&8&40&0.2 \
3&19&60&0.3167 \
4&32&84&0.3809 \
5&47&112&0.4196 \
6&64&144&0.4444 \
7&83&180&0.4611 \
end{array}
$$

To find the $x$-intercept, or the roots, set $y = 0$.

$$frac{x^2+6x-8x}{2x^2+10x+12} = 0$$

$$x^2+6x-8x = 0$$

$$x^2-2x = 0$$

$$x(x-2) = 0$$

$$x = 0 quad x = 2$$

To find the $y$-intercept, set $x = 0$.

$$frac{0^2+6(0)-8(0)}{2(0)^2+10(0)+12} = frac{0}{12} = 0$$

To find the vertical asymptote, see which value(s) of $x$ result in an undefined output. Set the denominator equal to $0$.

$$2x^2+10x+12 = 0$$

$$x^2+5x+6 = 0$$

$$(x+3)(x+2) = 0$$

$$x = -3 quad x = -2$$

To find the horizontal asymptote, find the limit to infinity. It becomes clear that as $x$ grows unboundedly large, the quadratic terms dominate. Both the numerator and denominator have leading quadratic terms, so the ratio of their coefficients will give the answer.

$$lim_{x to infty} frac{x^2+6x-8x}{2x^2+10x+12} = frac{x^2}{2x^2} = frac{1}{2}$$

Edit: If you meant $x^2+6x-8$ as the numerator (which you most probably did), you still carry out the same steps. For the $x$-intercepts, you reach

$$x^2+6x-8 = 0$$

which can’t be factored nicely, so just use the Quadratic Formula.

$$x = frac{-6pmsqrt{6^2-4(1)(-8)}}{2(1)} = frac{-6pmsqrt{68}}{2} = -3pmsqrt{17}$$

For the y-intercept, you get

$$frac{0^2+6(0)-8}{2(0)^2+10(0)+12} = frac{-8}{12} = -frac{2}{3}$$

The vertical and horizontal asympotes remain the same.