MATHWISE ACADEMY

Isolate $y$ in the linear equation:

$$y = x + 3 \quad \text{--- (Equation 1)}$$

Substitute Equation 1 into the quadratic equation:

$$x^2 + x(x + 3) = 14$$

$$x^2 + x^2 + 3x = 14$$

$$2x^2 + 3x - 14 = 0$$

Factorise the quadratic trinomial expression:

$$(2x + 7)(x - 2) = 0$$

Set each factor equal to zero to find $x$:

$$2x + 7 = 0 \implies x = -\frac{7}{2} \quad (-3.5)$$

$$x - 2 = 0 \implies x = 2$$

Substitute these $x$ values back into Equation 1 to find the corresponding $y$ values:

• If $x = 2 \implies y = 2 + 3 = 5$
• If $x = -3.5 \implies y = -3.5 + 3 = -0.5 \quad \left(-\frac{1}{2}\right)$

Solution Coordinates: $$(2;\ 5) \quad \text{and} \quad (-3.5;\ -0.5)$$

For an expression to be a perfect square, its corresponding equation must have equal roots ($\Delta = 0$):

$$\Delta = b^2 - 4ac = 0$$

Substitute $a = 1$, $b = -2k$, and $c = (3k + 4)$:

$$(-2k)^2 - 4(1)(3k + 4) = 0$$

$$4k^2 - 12k - 16 = 0$$

Divide the entire equation by 4:

$$k^2 - 3k - 4 = 0$$

Factorise the quadratic trinomial:

$$(k - 4)(k + 1) = 0$$

$$k = 4 \quad \text{or} \quad k = -1$$

To find $x$-intercepts, substitute $f(x) = 0$:

$$0 = -(x - 1)^2 + 9$$

$$(x - 1)^2 = 9$$

$$x - 1 = \pm 3$$

Separate the equations:

$$x - 1 = 3 \implies x = 4$$

$$x - 1 = -3 \implies x = -2$$

$$x\text{-intercepts} = (4;\ 0) \quad \text{and} \quad (-2;\ 0)$$

1. Calculate the $y$-intercept (substitute $x = 0$):

$$g(0) = \frac{6}{0 + 2} + 1 = \frac{6}{2} + 1 = 3 + 1 = 4$$

Coordinate: $$(0;\ 4)$$

2. Calculate the $x$-intercept (substitute $g(x) = 0$):

$$0 = \frac{6}{x + 2} + 1 \implies -1 = \frac{6}{x + 2}$$

$$-1(x + 2) = 6 \implies -x - 2 = 6 \implies -x = 8 \implies x = -8$$

Coordinate: $$(-8;\ 0)$$

Find where the parabola graph lies on or above the x-axis:

We calculated the horizontal x-intercepts of $f(x)$ to be $x = -2$ and $x = 4$ in Question 3.1.3.

Since the parabola is concave down ($a = -1$), the graph is above the x-axis ($f(x) \ge 0$) between the intercepts:

$$-2 \le x \le 4 \quad \text{or} \quad x \in [-2;\ 4]$$

For the hyperbola $g(x) = \frac{6}{x + 2} + 1$, the numerator is positive ($a = 6 > 0$).

This means the curves lie in the first and third quadrants relative to their asymptotes, and both curves are strictly decreasing throughout their domains.

The only point of exclusion is the vertical asymptote $x \neq -2$.

Decreasing for: $$x \in \mathbb{R}, \ x \neq -2 \quad \text{or} \quad x < -2 \ \cup \ x > -2$$

Identify the two coordinates:

• Turning point of $f(x) = (1;\ 9)$ (from Question 3.1.1)
• Asymptote intersection center of $g(x) = (-2;\ 1)$ (from Question 3.2.1)

Calculate gradient $m$ of $h(x)$:

$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 9}{-2 - 1} = \frac{-8}{-3} = \frac{8}{3}$$

Solve for equation using $y - y_1 = m(x - x_1)$ and coordinate point $(1; 9)$:

$$y - 9 = \frac{8}{3}(x - 1) \implies y = \frac{8}{3}x - \frac{8}{3} + 9$$

$$h(x) = \frac{8}{3}x + \frac{19}{3} \quad \text{or} \quad h(x) \approx 2.67x + 6.33$$

Analyze the equation $f(x) = k$:

$$-(x - 1)^2 + 9 = k \implies -(x - 1)^2 = k - 9 \implies (x - 1)^2 = 9 - k$$

For the equation to have no real roots, the term on the right must be strictly negative:

$$9 - k < 0 \implies 9 < k$$

$$k > 9$$

Alternative Graphical Method: Since the maximum range of the parabola $f(x)$ is $y \le 9$, the horizontal line $y = k$ will not intersect the parabola if $k > 9$. Thus, no real roots occur.