Analyze the following rational function for...? Best answer for correct answer

(x^4+7x^3-7x^2-43x+42)/(x^2+3x-4)
Zeroes
Y INT
MAxima and Minima
Increasing and Decreasing Intervals
End behavior (written in limit notation)
Continuity
Heavior at any discontinuity (written in limit notation

First, factor the numerator and the denominator.
The denominator factors as (x + 4)(x - 1).

Next, note that x = 1 is a zero of the numerator.
==> x^4 + 7x^3 - 7x^2 - 43x + 42 = (x - 1)(x^3 + 8x^2 + x - 42).

So, f(x) = (x - 1)(x^3 + 8x^2 + x - 42) / [(x + 4)(x - 1)].

**First of all, we have a hole (removable discontinuity) at x = 1.
The y-value there is given by (x^3 + 8x^2 + x - 42) / (x + 4) {at x = 1} = -32/5.
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Hereafter, we'll work with
f(x) = (x^3 + 8x^2 + x - 42)/(x + 4), with x ≠ 1.

(i) Zeros.
By the Rational Root Theorem, any rational zero of x^3 + 8x^2 + x - 42
must be a factor of -42.

By inspection, we find that x = 2 is a factor.
==> x^3 + 8x^2 + x - 42 = (x - 2)(x^2 + 10x + 21), by long/synthetic division
==> x^3 + 8x^2 + x - 42 = (x - 2)(x + 3)(x + 7).

Hence, the zeros (x-intercepts) are x = 2, -3, -7.
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(ii) y-intercept.
Setting x = 0 yields y = -42/4 = -21/2.
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(iii) Continuity issues.

Remember from above that we have a removable discontinuity/hole at (x, y) = (1, -32/5).

As for any other discontinuities:
By (i), we have f(x) = (x - 2)(x + 3)(x + 7) / (x + 4) with x ≠ 1.
==> We have an infinite discontinuity (vertical asymptote) at x = -4.
Moreover, lim(x→ -4+) f(x) = ∞, and lim(x→ -4-) f(x) = -∞.

In summary, f is continuous for all x ≠ -4, 1.
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(iv) Asymptotes.

From (iii), we know that x = -4 is a vertical asymptote.

As for others, first rewrite f after long division.
f(x) = (x^2 + 4x - 15) + 18/(x + 4).

As x→ ±∞, we have 18/(x + 4) → 0.
Hence, y = x^2 + 4x - 15 is a 'curved' asymptote(!).