Sketch the graph of $y = \frac{7-2x}{3x^2 + x - 14}$. Give the equations of the asymptotes and the coordinates of the turning points and the point(s) where the curve crosses the axes. Find the range of values of $k$ such that $\frac{7-2x}{3x^2 + x - 14} = k$ has only positive root(s).

1. Determine the equations of the asymptotes.

For the vertical asymptotes, set the denominator to $0$:
$3x^2 + x - 14 = 0$
$(3x + 7)(x - 2) = 0 \implies x = -\frac{7}{3} \text{ or } x = 2$

For the horizontal asymptote, evaluate the behaviour as $x \to \pm\infty$:
Dividing the numerator and denominator by $x^2$ gives:
$y = \frac{\frac{7}{x^2} - \frac{2}{x}}{3 + \frac{1}{x} - \frac{14}{x^2}}$
As $x \to \pm\infty$, $y \to \frac{0}{3} = 0$.
Hence, the horizontal asymptote is $y = 0$.

2. Determine the coordinates where the curve crosses the axes.

To find the $y$-intercept, substitute $x = 0$:
$y = \frac{7}{-14} = -\frac{1}{2}$

To find the $x$-intercept, set $y = 0$:
$7 - 2x = 0 \implies x = \frac{7}{2}$

The curve crosses the axes at $(0, -0.5)$ and $(3.5, 0)$.

3. Find the coordinates of the turning points using a graphing calculator.

By plotting $y = \frac{7-2x}{3x^2 + x - 14}$ on a graphing calculator and using the maximum/minimum features:
Local maximum point: $(0.542, -0.470)$ (to 3 s.f.)
Local minimum point: $(6.46, -0.0503)$ (to 3 s.f.)

4. Sketch the graph using the identified features.

Draw the asymptotes $x=2$, $x=-2.33$, and $y=0$ as dashed lines.
Plot the intercepts $(0, -0.5)$ and $(3.5, 0)$, and the turning points $(0.542, -0.470)$ and $(6.46, -0.0503)$.
Sketch the 3 branches of the curve:

• For $x < -\frac{7}{3}$, the curve is entirely above the $x$-axis, approaching $y=0$ as $x \to -\infty$ and going to $+\infty$ as $x \to -\frac{7}{3}^-$.
• For $-\frac{7}{3} < x < 2$, the curve comes from $-\infty$, passes through $(0, -0.5)$, reaches the local max at $(0.542, -0.470)$, and goes back down to $-\infty$ as $x \to 2^-$.
• For $x > 2$, the curve comes from $+\infty$, crosses the axis at $3.5$, reaches the local min at $(6.46, -0.0503)$, and approaches $y=0$ from below as $x \to \infty$.

1. Form a polynomial equation in $x$ and verify the linear case when $k=0$.

$\frac{7-2x}{3x^2+x-14} = k$

$3kx^2 + kx - 14k = 7 - 2x$

$3kx^2 + (k+2)x - (14k + 7) = 0$

Check the case when $k = 0$:
The equation becomes $(0+2)x - (0+7) = 0 \implies 2x = 7 \implies x = 3.5$.
Since $x = 3.5 > 0$, the equation has a single positive root. Thus, $k = 0$ is a valid value.

2. Find the condition for the quadratic equation to have real roots ($\Delta \ge 0$).

For $k \neq 0$, the equation is quadratic. For real roots, the discriminant must be non-negative ($\Delta \ge 0$):
$\Delta = (k+2)^2 - 4(3k)(-14k - 7)$
$\Delta = k^2 + 4k + 4 + 168k^2 + 84k$
$\Delta = 169k^2 + 88k + 4 \ge 0$

Find the critical values by solving $169k^2 + 88k + 4 = 0$:
$k = \frac{-88 \pm \sqrt{88^2 - 4(169)(4)}}{2(169)} = \frac{-88 \pm \sqrt{7744 - 2704}}{338}$
$k = \frac{-88 \pm \sqrt{5040}}{338} = \frac{-88 \pm 12\sqrt{35}}{338} = \frac{-44 \pm 6\sqrt{35}}{169}$

Thus, for real roots, we require:
$k \le \frac{-44-6\sqrt{35}}{169} \quad \text{or} \quad k \ge \frac{-44+6\sqrt{35}}{169}$

3. Apply conditions for both roots to be strictly positive.

For a quadratic equation to have only positive roots, both the sum of the roots and the product of the roots must be strictly positive.

Sum of roots $> 0$:
$-\frac{k+2}{3k} > 0 \implies \frac{k+2}{k} < 0 \implies -2 < k < 0$

Product of roots $> 0$:
$\frac{-(14k+7)}{3k} > 0 \implies \frac{2k+1}{k} < 0 \implies -0.5 < k < 0$

Taking the intersection of these two conditions, we require:
$-0.5 < k < 0$

4. Combine all conditions to determine the final exact range for $k$.

Take the intersection of the positive roots condition ($-0.5 < k < 0$) with the real roots condition:
$-0.5 < k \le \frac{-44-6\sqrt{35}}{169} \quad \text{or} \quad \frac{-44+6\sqrt{35}}{169} \le k < 0$

Finally, include the valid linear case ($k = 0$) to get the overall range:
$-0.5 < k \le \frac{-44-6\sqrt{35}}{169} \quad \text{or} \quad \frac{-44+6\sqrt{35}}{169} \le k \le 0$