Q. ( g(x) = dfrac{x^{2} + 5x – 14}{x^{2} + 4x – 21} ).

Problem

Given the rational function

\(g(x)=\dfrac{x^2+5x-14}{x^2+4x-21}\),

we will analyze it step by step: factor, simplify, determine domain, find any removable discontinuities (holes), vertical and horizontal asymptotes, and intercepts. Each step is shown with reasoning.

1. Factor numerator and denominator.

   Find two numbers that multiply to the constant term and add to the linear coefficient for each quadratic.

   Numerator: \(x^2+5x-14\). We seek a and b with \(ab=-14\) and \(a+b=5\). The pair \(7\) and \(-2\) works, so

   \(x^2+5x-14=(x+7)(x-2)\).

   Denominator: \(x^2+4x-21\). We seek a and b with \(ab=-21\) and \(a+b=4\). The pair \(7\) and \(-3\) works, so

   \(x^2+4x-21=(x+7)(x-3)\).

2. Write the factored form and simplify if possible.

   Substitute the factorizations into the original function:

   \(g(x)=\dfrac{(x+7)(x-2)}{(x+7)(x-3)}\).

   There is a common factor \((x+7)\) in numerator and denominator. Canceling it gives the simplified expression

   \(g_{\text{simp}}(x)=\dfrac{x-2}{x-3}\),

   but note: cancellation does not remove the restriction that made the canceled factor zero. We must record domain exclusions from the original denominator.

3. Determine the domain (values excluded).

   The original denominator is zero when \(x+7=0\) or \(x-3=0\), so

   \(x=-7\) and \(x=3\) are excluded from the domain.

   Domain: all real numbers except \(x=-7\) and \(x=3\).

4. Locate any removable discontinuity (hole).

   The common factor \((x+7)\) indicates a removable discontinuity at \(x=-7\). To find the y-coordinate of the hole, evaluate the simplified function at \(x=-7\):

   \(y=\dfrac{-7-2}{-7-3}=\dfrac{-9}{-10}=\dfrac{9}{10}.\)

   Therefore there is a hole at the point \(\left(-7,\dfrac{9}{10}\right).\)

5. Find vertical asymptotes.

   Vertical asymptotes occur at values where the original denominator is zero and the numerator is nonzero (or a nonremovable factor). We already have excluded \(x=3\) and \(x=-7\).

   At \(x=-7\) the factor was removable (cancelled), so that yields a hole, not an asymptote.

   At \(x=3\) the denominator is zero and the numerator is not zero (check: numerator at 3 is \(3^2+5\cdot3-14=9+15-14=10\neq0\)), so there is a vertical asymptote at

   \(x=3\).

6. Find horizontal asymptote.

   Compare degrees of numerator and denominator in the original function. Both are degree 2, so the horizontal asymptote is the ratio of the leading coefficients. Both leading coefficients are 1, so

   Horizontal asymptote: \(y=1\).

   Equivalently, the simplified form \(\dfrac{x-2}{x-3}\) also approaches 1 as \(|x|\to\infty\).

7. Find x- and y-intercepts.

   x-intercepts are solutions of the numerator equals zero (provided they are in the domain).

   Numerator zeros: from \((x+7)(x-2)=0\) we get \(x=-7\) and \(x=2\). But \(x=-7\) is excluded from the domain (it is a hole), so the only x-intercept is

   \((2,0)\).

   y-intercept: evaluate \(g(0)\) (0 is in the domain):

   \(g(0)=\dfrac{0^2+5\cdot0-14}{0^2+4\cdot0-21}=\dfrac{-14}{-21}=\dfrac{2}{3}.\)

   So the y-intercept is \(\left(0,\dfrac{2}{3}\right)\).

8. Summary (compact).

   Factored form: \(g(x)=\dfrac{(x+7)(x-2)}{(x+7)(x-3)}\).

   Simplified form (where defined): \(g_{\text{simp}}(x)=\dfrac{x-2}{x-3}\).

   Domain: all real numbers except \(x=-7\) and \(x=3\).

   Removable discontinuity (hole): \(\left(-7,\dfrac{9}{10}\right)\).

   Vertical asymptote: \(x=3\).

   Horizontal asymptote: \(y=1\).

   x-intercept: \((2,0)\).

   y-intercept: \(\left(0,\dfrac{2}{3}\right)\).