Alan

Anything multiplied by 0 is 0.

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rectangular form: x + iy

polar form: re^iθ

r = √(x² + y²)

θ = tan^-1(y/x)

$${\mathtt{r}} = {\sqrt{{{\mathtt{2}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{3}}}^{{\mathtt{2}}}}} \Rightarrow {\mathtt{r}} = {\mathtt{3.605\: \!551\: \!275\: \!463\: \!989\: \!3}}$$

$${\mathtt{theta}} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)} \Rightarrow {\mathtt{theta}} = {\mathtt{56.309\: \!932\: \!474\: \!02^{\circ}}}$$

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If x = 6, then y = (1/2)f(6/2) = (1/2)f(3) = (1/2)*4 = 2

so (6, 2) is on the graph of y = (1/2)f(x/2)

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Do you mean $$6.\overline{9}$$ ?

This means the 9 is repeated indefinitely: 6.999999.... etc, all the way to infinity.

It is equal to 7.

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See Chris's answer below.

An alternative method is to notice that the points are at the intersection of the straight line y = (1/2)x - 5 and a circle of radius 13 centred on (2, 7).

Circle equation:  (y - 7)² + (x - 2)² = 13²

Replace y by (1/2)x - 5

 (x/2- 5 - 7)² + (x - 2)² = 13²

Expand the brackets and simplify

5x²/4 - 16x + 148 = 169

Subtract 169 from both sides and multiply through the result by 4

5x² - 64x - 84 = 0

This factorises as

(5x + 6)(x - 14) = 0

so x = -6/5 (= -1.2) and x = 14

y = -(1/2)(6/5) - 5 = -28/5 = -5.6   and  y = (1/2)14 - 5 = 7 - 5 = 2

So the points are (-1.2, -5.6) and (14, 2)

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There are 4 values where the line y = -1 cuts f(x)

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(f(-4) - f(-1))/(-4 - (-1)) = (f(-2) - f(-1))/(-2 - (-1))

(4 - (-1))/(-4 + 1) = (f(-2) - (-1))/(-2 + 1)

(4 + 1)/(-3) = (f(-2) + 1)/(-1)

5/3 = f(-2) + 1

f(-2) = 5/3 - 1

f(-2) = 2/3

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Oops! Got sign wrong.  Thanks to Chris, I've now noticed this and corrected it.

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Geometric solution:

Triangle ADB is similar to triangle ABC so BD/AD = BC/AB ...(1)

Triangle BDC is similar to triangle ABC so CD/BD = BC/AB ...(2)

Multiply (1) and (2) together:  (BD/AD)*(CD/BD) = (BC/AB)²

or CD/AD = (3/1)² 

so CD/AD = 9/1

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Insert the values of x into either f(x) or g(x) and work out the values.

The points will be (1, f(1)); (-1, f(-1))

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