Detailed Answer

a) We have to find the positive root of the equation given. Let's first find the value of the discriminant:

\[\Delta = 5^{2} + \frac{4}{6} \cdot \frac{3}{2} = 25 + 1 = 26\]

Using this result into the quadratic formula:

\[ x = \frac{-5 - \sqrt{26}}{-\frac{2}{6}} = \frac{-5 - \sqrt{26}}{-\frac{1}{3}} = 15 + 3 \sqrt{26} \approx 30.3 \]

b) Writing the equation in vertex form we get the following:

\[ h(x) = -\frac{1}{6} x^{2} + 5x + \frac{3}{2} = -\frac{1}{6}\left(x^{2} - 30x + 15^{2}\right) + \frac{3}{2} + \frac{15^{2}}{6} = -\frac{1}{6}(x - 15)^{2} + 39 \]

As the \(y\)-coordinate of the vertex is 39, then the answer is 39 m.

c) Using the vertex form of the equation found in part (b), we can easily transform the function by stretching it vertically by \( \frac{1}{3} \) and horizontally by 2. Therefore:

\[ h_2(x) = \frac{1}{3} h\left(\frac{x}{2}\right) \] \[ = -\frac{1}{18}\left(\frac{x}{2} - 15\right)^{2} + \frac{39}{3} = -\frac{1}{18 \cdot 4}(x - 30)^{2} + 13 = -\frac{1}{72}(x - 30)^{2} + 13 \] \[ = -\frac{1}{72} x^{2} + \frac{5}{6} x + \frac{1}{2} \]

Detailed Answer

a) The function \( f(x) = 3 + 2 \cdot \cos(x) \) for \( x \in \mathbb{R} \) is translated by \( \binom{\alpha}{t} \), yielding the function \( g(x) = 3 + t + 2 \cdot \cos(x - \alpha) \). Evaluating \( g(x) \) at the given points, we get the following systems of equations:

\[ g(0) = 3 + t + 2 \cdot \cos(-\alpha) = 2 \] \[ g\left(\frac{\pi}{2}\right) = 3 + t + 2 \cdot \cos\left(\frac{\pi}{2} - \alpha\right) = 1 + \sqrt{3} \]

Using the fact that the cosine function is even and the identity \( \cos\left(\frac{\pi}{2} - \alpha\right) = \sin(\alpha) \), the above equations can be reduced to:

\[ t + 2 \cdot \cos(\alpha) = -1 \] \[ t + 2 \cdot \sin(\alpha) = -2 + \sqrt{3} \]

By eliminating \( t \), we get:

\[ 2 \cdot \cos(\alpha) - 2 \cdot \sin(\alpha) = 1 - \sqrt{3} \]

Squaring both sides of the equation and using the identities \( \cos^2(\alpha) + \sin^2(\alpha) = 1 \) and \( 2 \sin(\alpha) \cos(\alpha) = \sin(2\alpha) \), we obtain the equation:

\[ 2 - \sin(2\alpha) = \frac{2 - \sqrt{3}}{2} \]

Therefore, \( \sin(2\alpha) = \frac{\sqrt{3}}{2} \). The only values in the given range that satisfy this equation are \( \frac{\pi}{6} \) and \( \frac{\pi}{3} \).

Plugging in \( \frac{\pi}{6} \) into \( \cos(\alpha) - \sin(\alpha) \) gives \( \frac{\sqrt{3} - 1}{2} \), which is inconsistent with the above equations. Therefore, the solution is \( \frac{\pi}{3} \), and the value of \( t = -2 \).

Detailed Answer

a) We need to apply the transformation \( g(x) = f\left(x - \frac{2}{3}\right) + \frac{13}{6} \), so we can plug this into \( f(x) \). The transformation steps are as follows:

\[ g(x) = f\left(x - \frac{2}{3}\right) + \frac{13}{6} = \frac{ax - \frac{2a}{3} + b}{cx - \frac{2c}{3} + d} + \frac{13}{6} \]

\[ = \frac{6ax - 4a + 6b}{6cx - 4c + 6d} + \frac{13cx - \frac{26c}{3} + 13d}{6cx - 4c + 6d} \]

\[ = \frac{6ax + 13cx - 4a - \frac{26c}{3} + 6b + 13d}{6cx - 4c + 6d} \]

From this, we can write up a system of 4 equations, equating the respective coefficients:

\[ 6a + 13c = 231 \]

\[ -4a - \frac{26c}{3} + 6b + 13d = -409 \]

\[ 6c = 90 \]

\[ -4c + 6d = -186 \]

Solving this system gives the values \( a = 6 \), \( b = 3 \), \( c = 15 \), and \( d = 21 \).

b) Let's apply the transformations to \( f(x) = \frac{6x + 3}{15x - 21} \). We have:

\[ 5f\left(\frac{7}{3}x\right) = \frac{30\left(\frac{7}{3}x\right) + 15}{15\left(\frac{7}{3}x\right) - 21} = \frac{70x + 15}{35x - 21} \]

Next, we do polynomial division to obtain the required form:

\[ \frac{70x}{35x} = 2 \]

To find the remainder, we compute:

\[ 70x + 15 - 2(35x - 21) = 57 \]

Thus, we have the equation:

\[ h(x) = 2 + \frac{57}{35x - 21} \]

Therefore, the values are \( n = 2 \) and \( m = 57 \).

Detailed Answer

a) The x coordinate is mapped to \( \frac{x}{5} - 2 \) and the y coordinate is mapped to \( 3f(x) + 4 \), where \( f(x) = y \). Thus,

\[ g(x) = 3f\left(\frac{x}{5} - 2\right) + 4 \]

b) Firstly it has been vertically stretched by a factor 3, so \( q = 3 \). Then it was shifted in the y direction by \( b \), which here \( b = \frac{4}{3} \), as we need to divide by the vertical scale factor. Then it was stretched horizontally by \( p \), which is \( p = 5 \), as we have the \( \frac{1}{5} \) coefficient in front of the \( x \). Then it was translated horizontally by -2, so \( a = 2 \).

c) Here we can work backwards. We know that

\[ g(x) = 3f\left(\frac{x}{5} - 2\right) + 4 \]

So we can substitute \( g(x) = xe^x \), and rearrange for \( f(x) \).

\[ xe^x = 3f\left(\frac{x}{5} - 2\right) + 4 \]

\[ xe^x - 4 = 3f\left(\frac{x}{5} - 2\right) \]

\[ f\left(\frac{x}{5} - 2\right) = \frac{xe^x - 4}{3} \]

Here we can say \( u = \frac{x}{5} - 2 \) so \( x = 5(u + 2) \) which we can plug in:

\[ f(u) = \frac{5(u + 2)e^{5(u + 2)} - 4}{3} \]

\[ f(x) = \frac{5(x + 2)e^{5(x + 2)} - 4}{3} \]

Detailed Answer

a) We need to find the charge when \( t = 0 \).

\[ C(0) = 4900 - 4900 = 0\; \text{mAh} \]

b)

c) \( 60\% \) of 4900 is 2940, so we need to find when the charge reaches this level.

\[ 2940 = 4900 - 4900 \cdot 0.98^t \]

\[ 0.98^t = \frac{1960}{4900} = 0.4 \]

\[ t = \frac{\ln{0.4}}{\ln{0.98}} = 45\; \text{minutes} \]

d) The new max capacity is \( 1.1 \cdot 4900 = 5390 \). We also know the time needed to reach half capacity has been reduced by 22\%. First, we should calculate this time for the original model.

\[ \frac{4900}{2} = 4900 - 4900 \cdot 0.98^t \]

\[ 0.98^t = 0.5 \]

\[ t_{\frac{1}{2}} = \frac{\ln{0.5}}{\ln{0.98}} \]

The new time will be \( 0.78 \cdot t_{\frac{1}{2}} \). We then need to solve for the new decay rate of the battery, such that with this new time to half, the battery does indeed half. So,

\[ \frac{5390}{2} = 5390 \cdot r^{0.78 \cdot t_{\frac{1}{2}}} \]

Solving this for \( r \), in the same way as we did before, we get \( r = 0.97 \). So finally:

\[ C_2(t) = 5390 - 5390 \cdot 0.97^t \]

e) We simply need to equate \( C_2 \) to 4000.

\[ 4000 = 5390 - 5390 \cdot 0.97^t \]

\[ 0.97^t = \frac{1390}{5390} \]

\[ t = \frac{\ln{\frac{1390}{5390}}}{\ln{0.97}} = 44\; \text{minutes} \]

Detailed Answer

a) (-0.610, 0.972)

b) (i) (3.28, 13.7)

b) (ii) (-1.72, -12.7)

b) (iii)(-3.28, -13.7)

Detailed Answer

a) The function is translated 1 unit right and vertically stretched by a factor of 2.

b) On \( g(x) \), the coordinates are \( (0+1, 2 \cdot 5) \), so \( (1, 10) \). Then, on \( g(x) \), it has coordinates of \( (1+2, 10-1) \), resulting in \( (3, 9) \).

Detailed Answer

a) Factor a 3 out, then complete the square:

\[ f(x) = 3(x^2 - 8x + 17) = 3((x - 4)^2 - 16 + 17) = 3(x - 4)^2 + 1 \]

b) Now for \( g(x) \), we have:

\[ g(x) = 2f(x - 2) - 2 = 2(3(x - 6)^2 + 1) = 6(x - 6)^2 + 2 \]

Detailed Answer

a) \( 2(x + 2)^2 + 2 \)

b) \( (-2 + 2, 2 - 1) = (0, 1) \)

c) \( h(x) = 2(x + 2 - 2)^2 + 2 - 1 = 2x^2 + 1 \)

d) We need to find where the slope is equal to 6 for \( h(x) \), as that's where we can have a tangent with a slope of 6. The derivative is \( 4x \), thus \( 4x = 6 \), so \( x = 1.5 \).