Answer:
[tex]\textsf{y-intercept}: \quad \left(0,\dfrac{1}{4}\right)[/tex]
[tex]\textsf{x-intercepts}: \quad \left(-2+\sqrt{7},0\right) \textsf{ and }\left(-2-\sqrt{7},0\right)[/tex]
[tex]\textsf{vertical asymptote}: \quad x=-4[/tex]
[tex]\textsf{slant asymptote}: \quad y=-\dfrac{1}{3}x[/tex]
[tex]\textsf{domain}: \quad (- \infty, -4) \cup (-4, \infty)[/tex]
[tex]\textsf{range}: \quad (- \infty, \infty)[/tex]
Step-by-step explanation:
Given function:
[tex]f(x)=\dfrac{x^2+4x-3}{-3x-12}[/tex]
y-intercept
The y-intercept is the point at which the curve crosses the y-axis.
To find the y-intercept, substitute x = 0 into the function:
[tex]\begin{aligned} \implies f(0)& =\dfrac{(0)^2+4(0)-3}{-3(0)-12}\\\\ & =\dfrac{-3}{-12}\\\\ & = \dfrac{1}{4}\end{aligned}[/tex]
Therefore, the y-intercept is (0, 1/4)
x-intercepts
The x-intercepts are the points where the curve crosses the x-axis.
To find the x-intercepts, set the function to zero and solve for x:
[tex]\begin{aligned} f(x) & = 0\\\\\implies \dfrac{x^2+4x-3}{-3x-12} & = 0\\\\x^2+4x-3 & = 0\end{aligned}[/tex]
Solve using the quadratic formula:
[tex]x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0[/tex]
Therefore:
[tex]\implies x=\dfrac{-4 \pm \sqrt{4^2-4(1)(-3)}}{2(1)}[/tex]
[tex]\implies x=\dfrac{-4 \pm \sqrt{28}}{2}[/tex]
[tex]\implies x=\dfrac{-4 \pm 2\sqrt{7}}{2}[/tex]
[tex]\implies x=-2\pm\sqrt{7}[/tex]
Therefore, the x-intercepts are:
[tex]\left(-2+\sqrt{7},0\right) \textsf{ and }\left(-2-\sqrt{7},0\right)[/tex]
Vertical Asymptote
An asymptote is a line which the curve gets infinitely close to, but never touches.
The vertical asymptote is the value of x that makes the denominator of the function zero.
[tex]\implies -3x-12=0[/tex]
[tex]\implies x=4[/tex]
Therefore, the vertical asymptote is x = 4
Slant Asymptote
A slant asymptote occurs when the polynomial in the numerator of a rational function is a higher degree than the polynomial in the denominator.
To find the slant asymptote, divide the numerator by the denominator:
[tex]\large \begin{array}{r}-\frac{1}{3}x\phantom{)))))}\\-3x-12{\overline{\smash{\big)}\,x^2+4x-3\phantom{)}}}\\\underline{-~\phantom{(}(x^2+4x)\phantom{-))}}\\-3\phantom{)}\end{array}[/tex]
Therefore, the slant asymptote is:
[tex]y=-\dfrac{1}{3}x[/tex]
Domain
Input values (x-values): (-∞, -4) ∪ (-4, ∞)
Range
Output values (y-values): (-∞, ∞)
