20 points!!! F.4 Maths

Heidi

2009-10-27 4:12 am

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.

回答 (1)

Prof. Physics

2009-10-27 5:58 am

✔ 最佳答案

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.

參考: Physics king

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