On separate axes sketch the graphs of (i) $y = -3x + c$, where $c$ is a positive constant, (ii) $y = \frac{1}{x} + 5$ On each sketch show the coordinates of any point at which the graph crosses the $y$-axis and the equation of any horizontal asymptote - Edexcel - A-Level Maths Pure - Question 10 - 2017 - Paper 1

To sketch the graph of the equation $y = \frac{1}{x} + 5$:

• The graph has vertical asymptotes at $x = 0$ and a horizontal asymptote at $y = 5$.
• The graph will have two branches, one in the first quadrant approaching the asymptote and one in the third quadrant, showing the behavior on either side of the asymptotes.
• The $y$-intercept occurs at $(0, 5)$.

To prove that $(5 - c^2) > 12$, begin by substituting the graphical intersection points between $y = -3x + c$ and $y = \frac{1}{x} + 5$:

1. Set these equations equal to each other:
   $-3x + c = \frac{1}{x} + 5$
2. Rearranging gives:
   $\frac{1}{x} + 3x + 5 - c = 0$
3. Multiply through by $x$ (noting $x \neq 0$):
   $1 + 3x^2 + 5x - cx = 0$
4. The quadratic equation in the standard form is:
   $3x^2 + (5 - c)x + 1 = 0$
5. For there to be two distinct points of intersection, the discriminant must be greater than zero:
   $b^2 - 4ac > 0$ where $a = 3$, $b = (5 - c)$ and $c = 1$. This yields:
   $(5 - c)^2 - 4(3)(1) > 0$
6. Expanding gives:
   $(5 - c)^2 > 12$
7. This simplifies to:
   $5 - c^2 > 12$, confirming the desired inequality.

Using the derived inequality $(5 - c^2) > 12$, we can rewrite this as:

1. Rearranging gives:
   $-c^2 > 12 - 5$
   $-c^2 > 7$
2. Consequently, multiplying both sides by -1 (and flipping the inequality):
   $c^2 < -7$
3. Since there is no real solution for $c^2 < -7$, we go back to our original calculations to ensure:
   $c < 5 - \sqrt{12} ext{ and } c > 5 + \sqrt{12}$ This leads to the bounds for $c$. The possible range would be between $0 < c < 5 - \sqrt{12}$. Thus, evaluate the extremes to determine:
   $c < 5 - \sqrt{12} ext{ and } c > 5 + \sqrt{12}$.