Solutions in Linear-Quadratic Systems - Math 2

1. A quadratic function $h(x) = ax^2 + bx + c$ with $a > 0$ has its vertex below the $x$-axis. A linear function $\ell(x) = px + q$ intersects $h(x)$ at two distinct points. Which statement about the discriminant of the equation $h(x) = \ell(x)$ is always true?
   Answer: The discriminant is positive since there are two distinct intersection points When we solve $h(x) = \ell(x)$, we get $ax^2 + bx + c = px + q$, which rearranges to $ax^2 + (b-p)x + (c-q) = 0$. This is a quadratic equation whose solutions give the $x$-coordinates of intersection points. The discriminant of this equation is $\Delta = (b-p)^2 - 4a(c-q)$. Since the problem states that the functions intersect at two distinct points, the quadratic equation must have two distinct real solutions. This occurs if and only if the discriminant is positive: $\Delta > 0$. The fact that the vertex is below the $x$-axis is additional information about $h(x)$ itself, but doesn't directly determine the discriminant of the intersection equation. Choice A is incorrect because a zero discriminant would mean exactly one intersection point (tangency). Choice B is incorrect because the discriminant's sign depends on the specific functions, not just their types. Choice D is incorrect because the problem specifically states there are two distinct intersection points, which forces $\Delta > 0$.
2. Two functions are defined as $f(x) = x^2 - 6x + 9$ and $g(x) = mx + c$. If the graphs of $f(x)$ and $g(x)$ intersect at exactly one point, and this point has an $x$-coordinate of 3, what is the relationship between $m$ and $c$?
   Answer: $c = -3m$ First, note that $f(x) = x^2 - 6x + 9 = (x-3)^2$, so $f(x)$ is a parabola with vertex at $(3, 0)$. Since the functions intersect at exactly one point with $x$-coordinate 3, the line $g(x) = mx + c$ must be tangent to the parabola at $x = 3$. At $x = 3$: $f(3) = 0$, so the intersection point is $(3, 0)$. For tangency, the line must pass through $(3, 0)$: $g(3) = 3m + c = 0$, giving $c = -3m$. Additionally, for tangency, the derivatives must be equal at $x = 3$: $f'(x) = 2x - 6$, so $f'(3) = 0$. Since $g'(x) = m$, we need $m = 0$ for the slopes to match. If $m = 0$, then $c = 0$, and $g(x) = 0$ is a horizontal line through $(3, 0)$. This confirms tangency at the vertex. The relationship $c = -3m$ holds whether $m = 0$ or for the general case where we require the line to pass through the intersection point. Choice A gives $c = 3m$, which would mean $3m + c = 6m \neq 0$ unless $m = 0$. Choice C gives $c = 6m$, leading to $3m + c = 9m \neq 0$ unless $m = 0$. Choice D gives $c = -6m$, leading to $3m + c = -3m \neq 0$ unless $m = 0$. Only choice B ensures the line passes through the intersection point for any value of $m$.
3. Consider the system formed by $y = x^2 + px + q$ and $y = 2x + r$. If this system has exactly two solutions when $p = -2$ and $q = 1$, which condition on $r$ ensures the system has no real solutions?
   Answer: $r < -3$ With $p = -2$ and $q = 1$, the system becomes $x^2 - 2x + 1 = 2x + r$, which simplifies to $x^2 - 4x + (1-r) = 0$. The discriminant is $16 - 4(1-r) = 16 - 4 + 4r = 12 + 4r$. For no real solutions, we need discriminant < 0: $12 + 4r < 0$, so $4r < -12$, giving $r < -3$.
4. Consider all lines of the form $y = 2x + b$ where $b$ is a real number. How many of these lines intersect the parabola $y = x^2 - 4x + 5$ at exactly two distinct points?
   Answer: Infinitely many lines Setting the line equal to the parabola: $2x + b = x^2 - 4x + 5$, which rearranges to $x^2 - 6x + (5 - b) = 0$. For two distinct intersections, the discriminant must be positive: $36 - 4(5 - b) > 0$, so $36 - 20 + 4b > 0$, giving $16 + 4b > 0$, thus $b > -4$. Since $b$ can be any real number greater than $-4$, there are infinitely many such lines. Choice A would be correct if the discriminant were always non-positive. Choice B would apply if there were exactly one critical value. Choice C would be correct if only finitely many discrete values of $b$ worked.
5. A student claims that the system $\begin{cases} y = -x^2 + 4x - 3 \\ y = kx - 1 \end{cases}$ has exactly one solution when $k = 2$. To verify this claim, which approach would be most appropriate?
   Answer: Set the equations equal, solve the resulting quadratic, and verify that its discriminant equals zero To verify that a linear-quadratic system has exactly one solution, we must check that the discriminant of the resulting quadratic equation equals zero. Setting $-x^2 + 4x - 3 = 2x - 1$, we get $-x^2 + 2x - 2 = 0$, or $x^2 - 2x + 2 = 0$. The discriminant is $4 - 8 = -4 < 0$, meaning no real solutions, so the student's claim is false. Choice A is incorrect because a line passing through the vertex doesn't guarantee exactly one solution unless it's tangent. Choice C is imprecise for verification. Choice D only checks one specific point, which is insufficient to determine the total number of solutions.
6. The vertex of parabola $y = x^2 + 4x + k$ is at $(-2, k-4)$. For which value of $k$ will the horizontal line $y = 2$ intersect this parabola at exactly one point?
   Answer: $k = 6$ The parabola $y = x^2 + 4x + k$ can be written in vertex form. Completing the square: $y = (x+2)^2 - 4 + k = (x+2)^2 + (k-4)$. The vertex is at $(-2, k-4)$, which confirms the given information. Since the parabola opens upward, the vertex represents the minimum point. For the horizontal line $y = 2$ to intersect the parabola at exactly one point, the line must be tangent to the parabola at its vertex. This means the $y$-coordinate of the vertex must equal 2: $k - 4 = 2$, so $k = 6$. We can verify: when $k = 6$, the parabola becomes $y = x^2 + 4x + 6 = (x+2)^2 + 2$, with vertex at $(-2, 2)$. The line $y = 2$ passes through this vertex and, since it's horizontal and the vertex is the minimum point, intersects the parabola at exactly one point. Choice A gives vertex $y$-coordinate of $-2$, below the line. Choice B gives vertex $y$-coordinate of $0$, still below the line. Choice D gives vertex $y$-coordinate of $4$, above the line. Only choice C places the vertex exactly on the line $y = 2$.

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