dentify the vertical asymptotes of f(x) = quantity x plus 6 over quantity x squared minus 9x plus 18.

1 answer:

8 0

So the function we're dealing with is

f(x) = $\frac{x+6}{x^{2}-9x+18 }$

Factorising we get, $x^{2} -9x+18=0$
$(x-3)(x-6)=0$

Therefore, the vertical asymptotes are at x = 3 and x = 6

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1. Remember what we know about vertical angles and solve for x. (SHOW WORK)

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Answer:

Ans.1.) X = 7.

Ans.2.a.) AC =JL.

Ans.2.b.) BC = KL.

Step-by-step explanation:

For Question 1.

As we know vertically opposite angles are equal.

$\therefore (x+16)\° = (4x-5)\°\\\therefore (5+16)\° = (4x-x)\°\\ \therefore (3x) = (21)\\\therefore x = 7\\$

For Question 2.a)

$In\ \triangle ABC and \triangle JKL \\AB \cong JK\ \textrm{Given}\\\angle CAB \cong \angle LJK\ \textrm{each measure angle of 90\°}\\AC \cong JL\ \textrm{this is the required information to prove the triangles are congruent by SAS postulate}\\\therefore \triangle ABC \cong \triangle JKL\ \textrm{ By SAS postulate}$

For Question 2.b)

$In\ \triangle ABC and \triangle JKL \\AB \cong JK\ \textrm{Given}\\\angle CAB \cong \angle LJK\ \textrm{each measure angle of 90\°}\\BC \cong KL\ \textrm{this is the required information to prove the triangles are congruent by HL Theorem}\\\therefore \triangle ABC \cong \triangle JKL\ \textrm{ By Hypotenuse length Theorem}$