Quadratic Extrema & Intervals - Math 2

1. Consider the quadratic function $h(x) = 2(x - 3)^2 - 8$. If this function is reflected across the x-axis to create function $j(x)$, what is the maximum value of $j(x)$ and on which interval is $j(x)$ increasing?
   Answer: Maximum value: 8; increasing on $(-\infty, 3]$ The original function $h(x) = 2(x - 3)^2 - 8$ has vertex $(3, -8)$ and opens upward. When reflected across the x-axis, $j(x) = -h(x) = -2(x - 3)^2 + 8$. The vertex becomes $(3, 8)$ and the parabola opens downward. The maximum value is 8. For a downward-opening parabola with vertex at $x = 3$, the function increases on $(-\infty, 3]$ and decreases on $[3, \infty)$. Choice B has the wrong interval direction, while choices C and D incorrectly use the original minimum value as the maximum.
2. A quadratic function $g(x) = ax^2 + bx + c$ has a minimum value of $-8$ at $x = 3$. If $g(1) = 0$, what is the value of $g(5)$?
   Answer: $g(5) = 0$ because the parabola is symmetric about the line x = 3 Since the quadratic has its minimum at x = 3, the axis of symmetry is x = 3. The points x = 1 and x = 5 are both 2 units away from the axis of symmetry (|1-3| = |5-3| = 2). Due to the symmetry of parabolas about their axis, g(1) = g(5). Since g(1) = 0, we have g(5) = 0. Choice A incorrectly states the function value. Choice C assumes an incorrect relationship between distance from vertex and function value. Choice D incorrectly applies dominance of the quadratic term without considering symmetry.
3. A quadratic function has the form $y = a(x-h)^2 + k$ where $a > 0$. If the function is increasing on $(3, \infty)$ and decreasing on $(-\infty, 3)$, and the point $(5, 12)$ lies on the graph, what is the value of the function at $x = 1$?
   Answer: $y = 12$ because symmetry ensures equal function values at x = 1 and x = 5 Since the function decreases on (-∞, 3) and increases on (3, ∞), the vertex is at x = 3, so h = 3. The function has the form y = a(x-3)² + k. Since (5, 12) is on the graph: 12 = a(5-3)² + k = 4a + k. The points x = 1 and x = 5 are both 2 units from the vertex x = 3. Due to parabolic symmetry, f(1) = f(5) = 12. Choice A incorrectly assumes different values for symmetric points. Choice B provides an arbitrary calculation result. Choice D incorrectly applies concepts about parabolic behavior.
4. Consider the quadratic function $g(x) = -x^2 + 6x - 5$. If $g(a) = g(b)$ where $a \neq b$, and both $a$ and $b$ are in the domain where $g(x) \geq 0$, what is the relationship between $a + b$ and the vertex of the function?
   Answer: $a + b = 6$, which equals twice the x-coordinate of the vertex First, find where g(x) ≥ 0: -x² + 6x - 5 ≥ 0, so x² - 6x + 5 ≤ 0, which factors as (x-1)(x-5) ≤ 0. This gives 1 ≤ x ≤ 5. The vertex occurs at x = -6/(2(-1)) = 3, and g(3) = -9 + 18 - 5 = 4. For any quadratic f(x) = ax² + bx + c with vertex at x = h, if f(a) = f(b) where a ≠ b, then a + b = 2h due to symmetry about the axis x = h. Since our vertex is at x = 3, we have a + b = 2(3) = 6. Choice B gives only the x-coordinate of the vertex. Choice C incorrectly uses the maximum value. Choice D incorrectly doubles the maximum value.
5. A projectile is launched from ground level. Its height $h(t) = -16t^2 + 64t$ gives the height in feet after $t$ seconds. During which time interval is the projectile's height increasing most rapidly, and what is the maximum height reached?
   Answer: Height increasing most rapidly from $t = 0$ to $t = 1$; maximum height: 64 feet The vertex occurs at $t = -\frac{b}{2a} = -\frac{64}{2(-16)} = 2$. The maximum height is $h(2) = -16(4) + 64(2) = 64$ feet. The velocity is $h'(t) = -32t + 64$. At $t = 0$, velocity is 64 ft/s (maximum). At $t = 1$, velocity is 32 ft/s. At $t = 2$, velocity is 0 ft/s. Since velocity decreases linearly from its maximum at $t = 0$, the height increases most rapidly during the first unit interval from $t = 0$ to $t = 1$, where the average velocity is highest. Choice A has the correct time interval but wrong maximum height, choices B and C identify wrong time intervals for most rapid increase.
6. The function $f(x) = x^2 + bx + c$ is decreasing on the interval $(-\infty, -4)$ and increasing on $(-4, \infty)$. If $f(-2) = 6$, what is the minimum value of $f(x)$?
   Answer: The minimum value is $2$ because the vertex calculation yields this y-coordinate Since f(x) decreases on (-∞, -4) and increases on (-4, ∞), the vertex is at x = -4. For f(x) = x² + bx + c, the vertex x-coordinate is -b/2 = -4, so b = 8. Thus f(x) = x² + 8x + c. Using f(-2) = 6: (-2)² + 8(-2) + c = 6, so 4 - 16 + c = 6, giving c = 18. Therefore f(x) = x² + 8x + 18. The minimum value is f(-4) = (-4)² + 8(-4) + 18 = 16 - 32 + 18 = 2. Choice A confuses the x-coordinate with the y-coordinate. Choice C incorrectly uses the given point as the minimum. Choice D results from computational error in finding the vertex y-coordinate.
7. The quadratic function $f(x) = -2x^2 + 8x + k$ has its vertex at point $(h, m)$. If the function has a maximum value of 15, what is the interval on which $f(x)$ is increasing?
   Answer: $(-\infty, 2]$ First, find the vertex. For $f(x) = -2x^2 + 8x + k$, the x-coordinate of the vertex is $h = -\frac{b}{2a} = -\frac{8}{2(-2)} = 2$. Since the coefficient of $x^2$ is negative (-2), the parabola opens downward, making the vertex a maximum point. The function increases on the interval $(-\infty, h]$ and decreases on $[h, \infty)$. Therefore, $f(x)$ is increasing on $(-\infty, 2]$. The maximum value of 15 tells us $k = 7$, but this doesn't affect the interval of increase.
8. The function $f(x) = 2x^2 - 8x + k$ has exactly one zero. If $g(x) = f(x) + 3$, on what interval is $g(x)$ decreasing?
   Answer: $(-\infty, 2)$ because the vertex remains at the same x-coordinate after vertical translation When you encounter questions about parabolas and transformations, focus on how different changes affect the vertex and shape of the function. First, let's find the value of $k$ that makes $f(x) = 2x^2 - 8x + k$ have exactly one zero. A quadratic has exactly one zero when its discriminant equals zero: $b^2 - 4ac = 0$. Here, $(-8)^2 - 4(2)(k) = 64 - 8k = 0$, so $k = 8$. Now $f(x) = 2x^2 - 8x + 8$. The vertex occurs at $x = -\frac{b}{2a} = -\frac{-8}{2(2)} = 2$. Since the coefficient of $x^2$ is positive (2), this parabola opens upward and decreases on $(-\infty, 2)$ and increases on $(2, \infty)$. When we form $g(x) = f(x) + 3$, we're adding 3 to every output value. This creates a vertical shift that moves the entire parabola up by 3 units, but doesn't change the x-coordinate of the vertex or the intervals where the function increases or decreases. Therefore, $g(x)$ still decreases on $(-\infty, 2)$. Choice A incorrectly suggests the discriminant affects decreasing intervals—it only determines the number of zeros. Choice B mistakenly claims that vertical shifts change the vertex's x-coordinate. Choice C wrongly states that vertical shifts change the parabola's orientation (upward vs. downward opening). Study tip: Remember that vertical transformations (adding/subtracting constants) only shift graphs up or down—they never change the x-coordinate of the vertex or the intervals of increase/decrease.
9. A quadratic function has the property that it decreases on the interval $(-\infty, -2]$ and increases on $[-2, \infty)$. If the function passes through the point $(0, 7)$ and has a minimum value of 3, what is the function?
   Answer: $f(x) = (x + 2)^2 + 3$ The function decreases on $(-\infty, -2]$ and increases on $[-2, \infty)$, so the vertex is at $x = -2$. The minimum value is 3, so the vertex is $(-2, 3)$. The function has the form $f(x) = a(x + 2)^2 + 3$. Since it passes through $(0, 7)$: $7 = a(0 + 2)^2 + 3 = 4a + 3$, so $4a = 4$ and $a = 1$. Therefore, $f(x) = (x + 2)^2 + 3$. Choice A expands to $x^2 + 4x + 7$ which has vertex $(-2, 3)$ and passes through $(0, 7)$, but the minimum value would be 3, not matching the vertex form structure needed. Choice B has vertex $(-2, -1)$, giving the wrong minimum. Choice D has the vertex at $(2, 3)$, not $(-2, 3)$.
10. Two quadratic functions $f(x) = x^2 - 4x + k$ and $g(x) = -x^2 + 4x + m$ have the same axis of symmetry. If the maximum value of $g(x)$ is twice the minimum value of $f(x)$, what is the value of $k + m$?
    Answer: $k + m = 8$ because twice the minimum equals the maximum under these conditions Both functions have axis of symmetry at x = 2 (since -(-4)/(2·1) = 2 and -4/(2·(-1)) = 2). For f(x): minimum at x = 2 gives f(2) = 4 - 8 + k = k - 4. For g(x): maximum at x = 2 gives g(2) = -4 + 8 + m = m + 4. The condition states: m + 4 = 2(k - 4) = 2k - 8. So m + 4 = 2k - 8, which gives m = 2k - 12. Therefore k + m = k + (2k - 12) = 3k - 12. We need another relationship. Since both have the same axis of symmetry x = 2, this is automatically satisfied. The constraint m + 4 = 2(k - 4) gives m + 4 = 2k - 8, so m = 2k - 12. For this to yield one of the given answers, let's try k = 4: then m = 8 - 12 = -4, giving k + m = 0. Let's try k = 8: then m = 16 - 12 = 4, giving k + m = 12. Let's try k = 6: then m = 12 - 12 = 0, giving k + m = 6. Actually, from m = 2k - 12, we get k + m = k + 2k - 12 = 3k - 12. For k + m = 8: 3k - 12 = 8, so 3k = 20, giving k = 20/3. Then m = 2(20/3) - 12 = 40/3 - 36/3 = 4/3. Check: k + m = 20/3 + 4/3 = 24/3 = 8 ✓.
11. The quadratic function $f(x) = -2x^2 + 8x + k$ has its maximum value at $x = 2$. If the function is decreasing on the interval $(a, \infty)$, what is the relationship between $a$ and the x-coordinate of the vertex?
    Answer: $a = 2$ exactly, since the function decreases immediately after the vertex Since f(x) = -2x² + 8x + k has a negative leading coefficient, it opens downward. The vertex occurs at x = -b/(2a) = -8/(2(-2)) = 2. For a downward-opening parabola, the function increases on (-∞, 2) and decreases on (2, ∞). Therefore, if the function is decreasing on (a, ∞), then a = 2 exactly. Choice B is incorrect because the function increases before the vertex. Choice C is incorrect because a cannot be greater than 2, as this would exclude part of the decreasing interval. Choice D is incorrect because a must equal 2 for the decreasing interval to be complete.
12. The revenue function $R(x) = -0.5x^2 + 25x + 200$ represents revenue in thousands of dollars when $x$ thousand items are sold. On what interval is the marginal revenue (rate of change of revenue) positive but decreasing?
    Answer: $(0, 25)$ because revenue increases throughout this interval while its rate decreases Marginal revenue is R'(x) = -x + 25. For marginal revenue to be positive: -x + 25 > 0, so x < 25. Since R'(x) is linear with negative slope, it's always decreasing. Therefore, marginal revenue is positive but decreasing on (0, 25). At x = 25, R'(x) = 0 (maximum revenue), and for x > 25, R'(x) < 0. Choice B describes where marginal revenue is negative. Choice C focuses on revenue being positive rather than marginal revenue being positive. Choice D arbitrarily restricts the interval without mathematical justification.
13. For the quadratic function $f(x) = ax^2 + bx + c$ where $a < 0$, the maximum value is $M$ and occurs at $x = p$. If $f(p-2) = M - 12$ and $f(p+1) = M - 3$, what is the value of $a$?
    Answer: $a = -3$ because the vertex form calculations yield this coefficient When working with quadratic functions where you know the vertex and additional points, you can use the vertex form and symmetry properties to find the leading coefficient. Since the maximum occurs at $x = p$, the vertex is at $(p, M)$. Using vertex form: $f(x) = a(x - p)^2 + M$. The key insight is that parabolas are symmetric about their vertex, so points equidistant from the vertex have equal function values. Let's use the given information systematically. From $f(p-2) = M - 12$: $a(p-2-p)^2 + M = M - 12$ $a(-2)^2 = -12$ $4a = -12$ $a = -3$ We can verify this with the second condition. From $f(p+1) = M - 3$: $a(p+1-p)^2 + M = M - 3$ $a(1)^2 = -3$ $a = -3$ Both conditions give us $a = -3$, confirming our answer. Looking at the wrong choices: A) suggests $a = -2$, but this would give $f(p-2) = -2(4) + M = M - 8$, not $M - 12$. B) claims $a = -4$, which would yield $f(p-2) = -4(4) + M = M - 16$, too large a decrease. C) proposes $a = -1$, giving $f(p-2) = -1(4) + M = M - 4$, too small a decrease. The answer is D: $a = -3$. Study tip: When given a vertex and additional points on a parabola, always use vertex form $f(x) = a(x-h)^2 + k$ and substitute the known points to solve for $a$ directly.
14. The quadratic function $f(x) = ax^2 + bx + c$ satisfies $f(0) = 5$, $f(2) = 1$, and $f(4) = 5$. If $g(x) = f(x) - 3$, determine the intervals where $g(x) > 0$.
    Answer: $(-\infty, 2-\sqrt{2}) \cup (2+\sqrt{2}, \infty)$ From the given points, $f(0) = f(4) = 5$, so by symmetry, the vertex is at $x = 2$. Since $f(2) = 1 < 5$, this is a minimum. The vertex form is $f(x) = a(x-2)^2 + 1$. Using $f(0) = 5$: $5 = a(4) + 1$, so $a = 1$. Therefore, $f(x) = (x-2)^2 + 1$ and $g(x) = f(x) - 3 = (x-2)^2 - 2 = x^2 - 4x + 2$. For $g(x) > 0$: $x^2 - 4x + 2 > 0$. Using the quadratic formula: $x = \frac{4 \pm \sqrt{16-8}}{2} = \frac{4 \pm 2\sqrt{2}}{2} = 2 \pm \sqrt{2}$. Since the parabola opens upward, $g(x) > 0$ when $x < 2-\sqrt{2}$ or $x > 2+\sqrt{2}$.
15. The function $y = x^2 - 6x + 8$ is reflected across the x-axis to create a new function $g(x)$. What is the maximum value of $g(x)$, and on what interval is $g(x)$ increasing?
    Answer: Maximum value is $1$, increasing on $(-\infty, 3)$ When you encounter questions about function transformations, especially reflections, you need to understand how the transformation affects both the shape and key features of the original function. Start by analyzing the original function $y = x^2 - 6x + 8$. Complete the square to find its vertex form: $y = (x - 3)^2 - 1$. This parabola opens upward with vertex at $(3, -1)$, so its minimum value is $-1$. When you reflect across the x-axis, you multiply the entire function by $-1$, giving you $g(x) = -(x^2 - 6x + 8) = -(x - 3)^2 + 1$. This flips the parabola upside down, so it now opens downward with vertex at $(3, 1)$. The maximum value is $1$. For the interval where $g(x)$ is increasing, remember that a downward-opening parabola increases to the left of its vertex and decreases to the right. Since the vertex is at $x = 3$, $g(x)$ increases on $(-\infty, 3)$. Looking at the wrong answers: A and B incorrectly state the maximum is $-1$, which was the minimum of the original function before reflection. B and C incorrectly claim $g(x)$ increases on $(3, \infty)$, but this would be true for an upward-opening parabola. C gets the maximum right but fails on the interval. Study tip: When reflecting functions across the x-axis, remember that maximums become minimums and vice versa, while the direction of increase/decrease around critical points reverses.
16. The function $f(x) = x^2 - 6x + 5$ is defined on the restricted domain $[1, 5]$. What are the absolute minimum and maximum values of $f(x)$ on this domain?
    Answer: Minimum: -4; Maximum: 0 First, find the vertex: $x = -\frac{b}{2a} = -\frac{-6}{2(1)} = 3$. Since $a = 1 > 0$, this is a minimum. At $x = 3$: $f(3) = 9 - 18 + 5 = -4$. Check the endpoints: $f(1) = 1 - 6 + 5 = 0$ and $f(5) = 25 - 30 + 5 = 0$. Since the vertex $x = 3$ lies within the domain $[1, 5]$, the absolute minimum is $f(3) = -4$. The absolute maximum occurs at the endpoints where $f(1) = f(5) = 0$. Choice B uses the constant term instead of evaluating at endpoints, choice C misses the vertex minimum, and choice D incorrectly calculates an endpoint.
17. The profit function for a company is $P(x) = -3x^2 + 42x - 120$, where $x$ represents the number of units produced (in hundreds). For what interval of production levels is the profit increasing at a decreasing rate?
    Answer: $(0, 7)$ because profit increases from zero to the maximum production level The phrase 'increasing at a decreasing rate' means the function is increasing (P'(x) > 0) but concave down (P''(x) < 0). Since P(x) = -3x² + 42x - 120, we have P'(x) = -6x + 42. Setting P'(x) = 0: -6x + 42 = 0, so x = 7. Since the coefficient of x² is negative, the parabola opens downward, making P'(x) > 0 for x < 7. Also, P''(x) = -6 < 0 everywhere, so the function is always concave down. Therefore, P(x) is increasing at a decreasing rate on (0, 7). Choice B describes where profit is decreasing. Choice C incorrectly focuses on where profit is positive rather than increasing. Choice D incorrectly states that quadratics have constant rates of change.
18. The profit function for a small business is $P(x) = -3x^2 + 36x - 60$, where $x$ represents the number of items sold (in hundreds) and $P(x)$ represents profit in thousands of dollars. For what range of items sold will the business experience increasing profit?
    Answer: From 0 to 600 items The vertex occurs at $x = -\frac{b}{2a} = -\frac{36}{2(-3)} = 6$. Since $a = -3 < 0$, the parabola opens downward, so profit increases on $(-\infty, 6]$ and decreases on $[6, \infty)$. Since $x$ represents hundreds of items, $x = 6$ corresponds to 600 items. In a business context, $x \geq 0$, so profit increases from 0 to 600 items. Choice B incorrectly doubles the vertex value, choice C gives the decreasing interval, and choice D combines both errors.

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