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# Tutor profile: James A.

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James A.
Electronic Technician at Boeing and experienced STEM tutor
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## Questions

### Subject:Pre-Calculus

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Question:

Write the polar equation $$r = \cos\theta$$ in Cartesian form.

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James A.
Answer:

For converting between polar/Cartesian, we use 3 basic facts:$(x = r\cos\theta,$)$(y = r\sin\theta,$)$(x^2 + y^2 = r^2.$)We can change our polar equation into a Cartesian one using these facts. First, multipy $$r = \cos\theta$$ by $$r$$ to get:$(r^2 = r\cos\theta$)Now we can apply our facts to get:$(x^2 + y^2 = x$) This already solves our problem, but if we want, we can subtract $$x$$ from both sides and complete the square to get:$((x - \frac{1}{2})^2 + y^2 = \frac{1}{4},$)which is the equation of a circle with its center at $$(\frac{1}{2}, 0)$$ and a radius of $$\frac{1}{2}.$$

### Subject:Calculus

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Question:

What is the maximum value of: $$h(x) = -3x^2 + 6x - 1$$?

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James A.
Answer:

The extrema (max/mins) occur when $$h'(x) = 0$$. Since $$h'(x) = -6x + 6$$, that means that: $(-6x + 6 = 0,$)$(-6x = -6,$)$(x = 1.$)We can tell for sure that this is a maximum by the second derivative test:$(h''(1) = -6,$)$$-6 < 0$$, so that means that $$h$$ is concave down, and this is indeed a maximum. All that's left is to find the maximum value:$(h(1) = -3(1)^2 + 6(1) - 1 = -3 + 6 - 1 = 2$)

### Subject:Algebra

TutorMe
Question:

What is the maximum value of: $$h(x) = -3x^2 + 6x - 1$$ ?

Inactive
James A.
Answer:

$$h(x)$$ is a polynomial of degree 2, meaning 2 is the largest exponent of $$x$$. It's graph has the shape of a parabola. The parabola has a vertex at $$x = -\frac{b}{2a}$$, where $$b$$ is the coefficient of $$x$$ and $$a$$ is the coefficient of $$x^2$$. For $$h(x)$$, this means that the vertex is at $$x = -\frac{6}{2 \times -3} = -\frac{6}{-6} = 1$$. But the maximum value refers to the value of $$h$$, not $$x$$. To find that, we need $$h(1)$$, which is: $(h(1) = -3(1)^2 + 6(1) - 1 = -3 + 6 - 1 = 2$)

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