Question: 1 Find the values of $k$ for which the line $y = kx + 1$ is a tangent to the curve $y = x^2 - 7x + 2$.
1. **State the problem:** Find values of $k$ such that the line $y = kx + 1$ is tangent to the curve $y = x^2 - 7x + 2$.
2. **Formula and rules:** For tangency, the line and curve must intersect at exactly one point. Set the line equal to the curve:
$$kx + 1 = x^2 - 7x + 2$$
Rearranged:
$$x^2 - 7x - kx + 2 - 1 = 0 \implies x^2 - (7 + k)x + 1 = 0$$
3. **Condition for tangency:** The quadratic equation in $x$ must have exactly one solution, so its discriminant $D$ must be zero:
$$D = b^2 - 4ac = 0$$
Here, $a = 1$, $b = -(7 + k)$, $c = 1$.
4. **Calculate discriminant:**
$$D = (-(7 + k))^2 - 4(1)(1) = (7 + k)^2 - 4 = 0$$
5. **Solve for $k$:**
$$ (7 + k)^2 = 4 $$
$$ 7 + k = \pm 2 $$
6. **Find values:**
- If $7 + k = 2$, then $k = 2 - 7 = -5$
- If $7 + k = -2$, then $k = -2 - 7 = -9$
**Final answer:**
$$k = -5 \text{ or } k = -9$$