標題:
20 points!!! F.4 Maths
發問:
1. The vertex of the graph of a quadratic function y=f(x) is (4,-3), and the graph passes through the point (1,15). Find f(x).2. Find the x-intercepts of the graph.a) y = 2x^2+5x+3b) y = -x^2+14x-493. Given that f(x) is a function of x where f(1+x) = f(1)．f(x) and f(1) ≠ 0, show thata) f(0) =... 顯示更多 1. The vertex of the graph of a quadratic function y=f(x) is (4,-3), and the graph passes through the point (1,15). Find f(x). 2. Find the x-intercepts of the graph. a) y = 2x^2+5x+3 b) y = -x^2+14x-49 3. Given that f(x) is a function of x where f(1+x) = f(1)．f(x) and f(1) ≠ 0, show that a) f(0) = 1 b) f(-1) = 1/f(1) 4. A function is called an odd function if f(-x) = -f(x). Show that f(x) = x^3-3x is an odd funcion.
1. A quadratic equation can be written as y = a(x - h)^2 + k Where (h , k) is the vertex So, h = 4, k = -3 y = a(x - 4)^2 - 3 Put (x , y) = (1 , 15) 15 = a(1 - 4)^2 - 3 a = 2 So, the required equation: y = 2(x - 4)^2 - 3 y = 2(x^2 - 8x + 16) - 3 y = 2x^2 - 16x + 29 2. Put y = 0 to find the x-intercept. a. 0 = 2x^2 + 5x + 3 (2x + 3)(x + 1) = 0 x = -1 or -3/2 So, the x-intercepts are x = -1 and x = -3/2 b. 0 = -x^2 + 14x - 49 x^2 - 14x + 49 = 0 (x - 7)^2 = 0 x = 7 The x-intercept is x = 7 3.a. Put x = 0 f(1 + 0) = f(1)?f(0) f(1) = f(1)?f(0) f(0) = 1 b. Put x = -1 f(1 - 1) = f(1)?f(-1) f(0) = f(1)?f(-1) 1 = f(1)?f(-1) f(-1) = 1/f(1) 4. f(-x) = (-x)^3 - 3(-x) = -x^3 + 3x = -(x^3 - 3x) = -f(x) So, f(x) is odd.
其他解答:
20 points!!! F.4 Maths
發問:
1. The vertex of the graph of a quadratic function y=f(x) is (4,-3), and the graph passes through the point (1,15). Find f(x).2. Find the x-intercepts of the graph.a) y = 2x^2+5x+3b) y = -x^2+14x-493. Given that f(x) is a function of x where f(1+x) = f(1)．f(x) and f(1) ≠ 0, show thata) f(0) =... 顯示更多 1. The vertex of the graph of a quadratic function y=f(x) is (4,-3), and the graph passes through the point (1,15). Find f(x). 2. Find the x-intercepts of the graph. a) y = 2x^2+5x+3 b) y = -x^2+14x-49 3. Given that f(x) is a function of x where f(1+x) = f(1)．f(x) and f(1) ≠ 0, show that a) f(0) = 1 b) f(-1) = 1/f(1) 4. A function is called an odd function if f(-x) = -f(x). Show that f(x) = x^3-3x is an odd funcion.
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最佳解答:1. A quadratic equation can be written as y = a(x - h)^2 + k Where (h , k) is the vertex So, h = 4, k = -3 y = a(x - 4)^2 - 3 Put (x , y) = (1 , 15) 15 = a(1 - 4)^2 - 3 a = 2 So, the required equation: y = 2(x - 4)^2 - 3 y = 2(x^2 - 8x + 16) - 3 y = 2x^2 - 16x + 29 2. Put y = 0 to find the x-intercept. a. 0 = 2x^2 + 5x + 3 (2x + 3)(x + 1) = 0 x = -1 or -3/2 So, the x-intercepts are x = -1 and x = -3/2 b. 0 = -x^2 + 14x - 49 x^2 - 14x + 49 = 0 (x - 7)^2 = 0 x = 7 The x-intercept is x = 7 3.a. Put x = 0 f(1 + 0) = f(1)?f(0) f(1) = f(1)?f(0) f(0) = 1 b. Put x = -1 f(1 - 1) = f(1)?f(-1) f(0) = f(1)?f(-1) 1 = f(1)?f(-1) f(-1) = 1/f(1) 4. f(-x) = (-x)^3 - 3(-x) = -x^3 + 3x = -(x^3 - 3x) = -f(x) So, f(x) is odd.
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