# Tutor profile: Cera H.

Inactive

Cera H.

Electrical Engineer and Math Tutor

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## Questions

### Subject: Pre-Calculus

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Question:

Find the vertical and horizontal asymptotes: $$ f(x) = \frac{4x^2}{x^2 - 4} $$

Inactive

Cera H.

Answer:

Vertical: $$ x^2 - 4 = 0 \Rightarrow x = \pm2 $$ Horizontal: Since degrees in numerator and denominator are the same, the horizontal asymptote is the ratio of leading coefficients, so $$ y = \frac{4}{1} = 4 $$ Vertical: $$ x = \pm2 $$ , Horizontal: y = 4

### Subject: Trigonometry

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Question:

Prove $$ tan^2x + 1 = sec^2x $$

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Cera H.

Answer:

$$ tan^2x + 1 = \frac{sin^2x}{cos^2x} + \frac{cos^2x}{cos^2x} = \frac{1}{cox^2x} = sec^2x $$

### Subject: Algebra

TutorMe

Question:

If x = -4, what is the value of $$ \frac{x^2 - 1}{x + 1} $$?

Inactive

Cera H.

Answer:

$$ \frac{(-4)^2 - 1}{(-4) + 1} = \frac{16 - 1}{-3} = \frac{15}{-3} = -5 $$

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