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# 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.

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.

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

### Subject:Algebra

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Question:

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

Inactive
Cera H.

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

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