Question: Find the value of $m$ for which $y = mx + 3$ is tangent to $y = x^2 + 5x + 7$.

Step 1. Tangent means line and curve meet at exactly ONE point (repeated root).

Substitute: $mx + 3 = x^2 + 5x + 7$:

$$x^2 + (5 - m)x + 4 = 0$$

Step 2. For tangency, discriminant $= 0$:

$$(5 - m)^2 - 16 = 0$$ $$(5 - m)^2 = 16$$ $$5 - m = \pm 4$$ $$m = 1 \text{ or } m = 9$$

Answer: $m = 1$ or $m = 9$ (two tangent lines exist, one on each side).

Tip: "Line is tangent to curve" → substitute, form a quadratic in $x$, impose discriminant $= 0$. Often two values of the parameter give tangency — both are valid unless the question restricts.

Question: Find the values of $k$ for which $y = kx - 1$ meets $y = x^2 + 3$ at two distinct points.

Step 1. Two distinct intersections means discriminant $> 0$ (strict).

Substitute: $kx - 1 = x^2 + 3 \Rightarrow x^2 - kx + 4 = 0$.

Step 2. Discriminant condition:

$$k^2 - 16 > 0$$ $$(k-4)(k+4) > 0$$ $$k < -4 \text{ or } k > 4$$

Answer: $k < -4$ or $k > 4$.

Tip: "Two distinct intersections" $\Rightarrow$ discriminant $> 0$. Solve the quadratic inequality in the parameter. The sign analysis follows from the parabola direction.

Question: Find the values of $c$ for which $y = 2x + c$ never meets $y = x^2 - 4x + 5$.

Step 1. "Never meets" means substituted quadratic has no real roots: discriminant $< 0$.

$2x + c = x^2 - 4x + 5 \Rightarrow x^2 - 6x + (5 - c) = 0$.

Step 2. Discriminant $< 0$:

$$36 - 4(5 - c) < 0$$ $$36 - 20 + 4c < 0$$ $$16 + 4c < 0$$ $$c < -4$$

Answer: $c < -4$.

Tip: "Never intersects" $\Rightarrow$ discriminant $< 0$ (strict). Solve as a linear or quadratic inequality in the parameter.

Question: Show that the equation $(k+1)x^2 + 2x + k = 0$ has real roots for all real $k$ (where $k \neq -1$).

Step 1. Compute discriminant:

$$\Delta = 4 - 4(k+1)k = 4 - 4k^2 - 4k = 4(1 - k^2 - k)$$

Wait — that's not always $\geq 0$. Let me re-state: we want to determine the range of $k$ for which roots are real.

Step 1'. $\Delta \geq 0$:

$$1 - k^2 - k \geq 0$$ $$k^2 + k - 1 \leq 0$$

Step 2. Roots of $k^2 + k - 1 = 0$: $k = \dfrac{-1 \pm \sqrt{5}}{2}$.

So $k^2 + k - 1 \leq 0$ when $\dfrac{-1 - \sqrt{5}}{2} \leq k \leq \dfrac{-1 + \sqrt{5}}{2}$, i.e. approximately $-1.618 \leq k \leq 0.618$.

Answer: Real roots when $k \in \left[\dfrac{-1-\sqrt{5}}{2}, \dfrac{-1+\sqrt{5}}{2}\right]$, excluding $k = -1$.

Tip: Parametric discriminant problems: $\Delta(k) \geq 0$ (real) or $> 0$ (distinct). The resulting condition on $k$ is itself usually a polynomial inequality.

Question: Show that $x^2 - 6x + 11 > 0$ for all real $x$.

Step 1. Two ingredients:

• Leading coefficient of $x^2$ is $+1 > 0$ (parabola opens upwards).
• Discriminant $= 36 - 44 = -8 < 0$ (no real roots — parabola doesn't touch the $x$-axis).

Step 2. Combine: the parabola opens upwards AND lies entirely above the $x$-axis. Therefore:

$$x^2 - 6x + 11 > 0 \text{ for all real } x \;\checkmark$$

Alternative — completing the square:

$$x^2 - 6x + 11 = (x - 3)^2 + 2 \geq 2 > 0$$

Tip: To show "$f(x) > 0$ always":

1. Show leading coefficient $> 0$ (or complete the square with positive constant).
2. Show discriminant $< 0$.

Both must hold for STRICT inequality. For $\geq 0$, allow discriminant $\leq 0$ (touching the axis is OK).

Question: Express $y = 2x^2 - 12x + 23$ in the form $a(x - h)^2 + k$. State the minimum value of $y$ and the value of $x$ at which it occurs.

Step 1. Factor out the leading coefficient from the $x$-terms:

$$y = 2(x^2 - 6x) + 23$$

Step 2. Complete the square inside the bracket:

$$x^2 - 6x = (x - 3)^2 - 9$$

Step 3. Substitute back:

$$y = 2[(x - 3)^2 - 9] + 23 = 2(x - 3)^2 - 18 + 23 = 2(x - 3)^2 + 5$$

So $a = 2$, $h = 3$, $k = 5$.

Step 4. Since $2(x - 3)^2 \geq 0$ for all $x$ (with equality at $x = 3$):

$$y \geq 5$$

Minimum value of $y = 5$, occurring at $x = 3$.

Tip: Vertex form $a(x - h)^2 + k$:

• If $a > 0$: min value $= k$ at $x = h$.
• If $a < 0$: max value $= k$ at $x = h$.

For completing the square, divide the $x$-coefficient by $2$, square it, add and subtract.

Question: The equation $kx^2 + (k+1)x + k = 0$ has real roots. Find the range of $k$. (Treat $k$ as a real parameter, allowing $k = 0$ as the degenerate linear case.)

Step 1. Two cases:

Case 1: $k = 0$. Equation becomes $x = 0$. Real root exists.

Case 2: $k \neq 0$. Quadratic with real roots: discriminant $\geq 0$.

$$\Delta = (k+1)^2 - 4k^2 \geq 0$$ $$k^2 + 2k + 1 - 4k^2 \geq 0$$ $$-3k^2 + 2k + 1 \geq 0$$ $$3k^2 - 2k - 1 \leq 0$$ $$(3k + 1)(k - 1) \leq 0$$

Step 2. Sign analysis: $(3k+1)(k-1) \leq 0$ when $-\dfrac{1}{3} \leq k \leq 1$.

Step 3. Combine cases: $k = 0$ (covered by Case 1) and $k \in \left[-\dfrac{1}{3}, 1\right] \setminus \{0\}$ (Case 2). Total: $k \in \left[-\dfrac{1}{3}, 1\right]$.

Answer: $k \in \left[-\dfrac{1}{3}, 1\right]$.

Tip: Multi-condition problems often need:

1. Special case (e.g. $k = 0$ degenerates the quadratic).
2. Main case with discriminant condition.
3. Combine the answer sets.