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Comparing Quadratic Functions - Math 2

  1. Function p(x)=ax2+bx+cp(x) = ax^2 + bx + c passes through points (−1,8)(-1, 8), (0,3)(0, 3), and (2,7)(2, 7). Function q(x)=2x2−x+1q(x) = 2x^2 - x + 1. On the interval [1,3][1, 3], which function grows faster?
    Answer: Function qq grows faster because its average rate of change is 7 First, find p(x) using the three points. From (0,3): c = 3. Using (-1,8) and (2,7): a - b + 3 = 8 and 4a + 2b + 3 = 7, giving a - b = 5 and 4a + 2b = 4, so 2a + b = 2. Solving: a = 1, b = 0. Thus p(x) = x² + 3. On [1,3]: p(1) = 4, p(3) = 12, average rate = (12-4)/(3-1) = 4. For q(x): q(1) = 2(1) - 1 + 1 = 2, q(3) = 2(9) - 3 + 1 = 16, average rate = (16-2)/(3-1) = 7. Since 7 > 4, function q grows faster.
  2. Two quadratic functions u(x)=−x2+6x−5u(x) = -x^2 + 6x - 5 and v(x)=2x2−8x+11v(x) = 2x^2 - 8x + 11 are compared on the interval [1,4][1, 4]. Which function has the greater average rate of change, and why?
    Answer: Function vv because its rate of change is 22 compared to uu's rate of 11 Calculate the average rates of change on [1,4]. For u(x): u(1) = -1 + 6 - 5 = 0, u(4) = -16 + 24 - 5 = 3. Average rate = (3-0)/(4-1) = 1. For v(x): v(1) = 2 - 8 + 11 = 5, v(4) = 32 - 32 + 11 = 11. Average rate = (11-5)/(4-1) = 2. Since 2 > 1, function v has the greater average rate of change.
  3. Consider the quadratic functions F(x)=−2x2+16x−30F(x) = -2x^2 + 16x - 30 and G(x)=x2−6x+13G(x) = x^2 - 6x + 13. On the interval [3,6][3, 6], which function has the greater average rate of change?
    Answer: Function GG because its average rate of change is 33 Calculate average rates on [3,6]. For F(x): F(3) = -18 + 48 - 30 = 0, F(6) = -72 + 96 - 30 = -6. Average rate = (-6-0)/(6-3) = -2. For G(x): G(3) = 9 - 18 + 13 = 4, G(6) = 36 - 36 + 13 = 13. Average rate = (13-4)/(6-3) = 3. Since 3 > -2, function G has the greater average rate of change. F decreases on this interval (vertex at x = 4), while G increases throughout (vertex at x = 3).
  4. Function r(x)=x2−2x+4r(x) = x^2 - 2x + 4 and function s(x)=−0.5x2+4x−3s(x) = -0.5x^2 + 4x - 3 are analyzed on the interval [0,2][0, 2]. Based on their average rates of change, which conclusion is most accurate?
    Answer: Function ss grows faster with an average rate of change of 33 Calculate average rates on [0,2]. For r(x): r(0) = 4, r(2) = 4 - 4 + 4 = 4. Average rate = (4-4)/(2-0) = 0. For s(x): s(0) = -3, s(2) = -0.5(4) + 4(2) - 3 = -2 + 8 - 3 = 3. Average rate = (3-(-3))/(2-0) = 3. Since 3 > 0, function s grows faster with an average rate of change of 3.
  5. Function A(x)=3x2−12x+15A(x) = 3x^2 - 12x + 15 and function B(x)=−x2+8x−12B(x) = -x^2 + 8x - 12 are compared on interval [2,5][2, 5]. Which function exhibits faster growth over this interval?
    Answer: Function AA because its average rate of change is 99 Calculate average rates on [2,5]. For A(x): A(2) = 12 - 24 + 15 = 3, A(5) = 75 - 60 + 15 = 30. Average rate = (30-3)/(5-2) = 9. For B(x): B(2) = -4 + 16 - 12 = 0, B(5) = -25 + 40 - 12 = 3. Average rate = (3-0)/(5-2) = 1. Function A grows much faster with rate 9 vs B's rate of 1. The vertex of A is at x = 2 and it increases from there. The vertex of B is at x = 4, so B increases then decreases on [2,5].
  6. Two quadratic functions are given: h(x)=x2−4x+7h(x) = x^2 - 4x + 7 and k(x)=3(x−1)2+2k(x) = 3(x-1)^2 + 2. Over which interval does function kk have a greater average rate of change than function hh?
    Answer: [1,3][1, 3] because kk increases more rapidly away from its vertex First, find the vertices: h(x) has vertex at x = 2, k(x) has vertex at x = 1. For [1,3]: h(1) = 4, h(3) = 4, so average rate = 0. k(1) = 2, k(3) = 14, so average rate = 6. For [0,1]: h(0) = 7, h(1) = 4, rate = -3. k(0) = 5, k(1) = 2, rate = -3. For [2,4]: h(2) = 3, h(4) = 7, rate = 2. k(2) = 5, k(4) = 29, rate = 12. For [-1,1]: h(-1) = 12, h(1) = 4, rate = -4. k(-1) = 14, k(1) = 2, rate = -6. Only on [1,3] does k have a greater (more positive) average rate of change.
  7. Two quadratic functions p(x)=ax2+bx+cp(x) = ax^2 + bx + c and q(x)=dx2+ex+fq(x) = dx^2 + ex + f satisfy the following: p(1)=3p(1) = 3, p(3)=7p(3) = 7, p(5)=15p(5) = 15, and q(1)=8q(1) = 8, q(3)=2q(3) = 2, q(5)=4q(5) = 4. Over the interval [1,5][1, 5], which statement is most accurate?
    Answer: Function pp has a greater average rate of change than qq, with pp accelerating and qq decelerating For p(x): average rate over [1,5] = (15-3)/(5-1) = 3. The differences are: p(3)-p(1) = 4, p(5)-p(3) = 8, showing increasing rate (acceleration). For q(x): average rate over [1,5] = (4-8)/(5-1) = -1. The differences are: q(3)-q(1) = -6, q(5)-q(3) = 2, showing the rate is becoming less negative (deceleration in the negative direction). Since 3 > -1, function p has greater average rate of change. Function p shows acceleration (increasing positive rate), while q shows deceleration (rate changing from negative to positive). Choice A is wrong about q having consistently decreasing rate. Choice B is wrong because p has greater average rate. Choice C is wrong because average rates differ (3 vs -1).
  8. Function m(x)=0.5x2+3x−1m(x) = 0.5x^2 + 3x - 1 and function n(x)n(x) has the property that n(0)=5n(0) = 5, n(2)=1n(2) = 1, and n(4)=5n(4) = 5. Over the interval [0,4][0, 4], which statement about their growth patterns is correct?
    Answer: Function mm grows faster because its total change is 20 units For m(x): m(0) = -1, m(4) = 0.5(16) + 3(4) - 1 = 8 + 12 - 1 = 19. Total change = 19 - (-1) = 20 units. For n(x): n(0) = 5, n(4) = 5. Total change = 0 units. Average rate for m: 20/4 = 5. Average rate for n: 0/4 = 0. Function m grows faster with a positive average rate of change compared to n's zero average rate.
  9. Consider the quadratic functions h(x)=2x2−8x+11h(x) = 2x^2 - 8x + 11 and j(x)=−x2+6x−4j(x) = -x^2 + 6x - 4. Over the interval [2,4][2, 4], which statement about their comparative behavior is most accurate?
    Answer: Function hh is increasing while jj is decreasing, making direct rate comparison challenging but meaningful For h(x): vertex at x = 2, so h is increasing for x > 2. Over [2,4]: h(2) = 8-16+11 = 3, h(4) = 32-32+11 = 11, so h increases by 8. For j(x): vertex at x = 3, so j increases for x < 3 and decreases for x > 3. j(2) = -4+12-4 = 4, j(3) = -9+18-4 = 5, j(4) = -16+24-4 = 4. So j increases from 2 to 3, then decreases from 3 to 4, with net change = 0. Since h is consistently increasing while j increases then decreases over [2,4], choice B correctly identifies this behavior. The comparison is meaningful because we can compare average rates (8/2 = 4 for h vs 0/2 = 0 for j). Choice A is wrong because j isn't consistently increasing. Choice C has the directions reversed. Choice D is wrong because h isn't decreasing.
  10. Two quadratic functions f(x)=2x2−4x+1f(x) = 2x^2 - 4x + 1 and g(x)=−x2+4x−1g(x) = -x^2 + 4x - 1 are analyzed over the interval [0,3][0, 3]. Which statement best characterizes how their growth patterns compare?
    Answer: Function gg increases then decreases symmetrically, while ff decreases then increases with greater final slope For f(x): vertex at x = 1, f'(x) = 4x - 4. f decreases for x < 1, increases for x > 1. f(0) = 1, f(1) = -1, f(3) = 7. For g(x): vertex at x = 2, g'(x) = -2x + 4. g increases for x < 2, decreases for x > 2. g(0) = -1, g(2) = 3, g(3) = 2. Over [0,3]: f decreases then increases (1 to -1 to 7), while g increases then decreases (-1 to 3 to 2). Function g's behavior is more symmetric around its vertex at x = 2. Function f has steeper final slope: f'(3) = 8, while g'(3) = -2. Choice C correctly describes both behaviors and notes f's greater final slope magnitude.

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