Question

Функция задана формулой f(x)=x^2-16 a) найдите f(-2) , f(0)б) найдите значения х, при которых:f(х)=-12 f(x)=0функция задана формулой y=-2x+1б)найдите координаты точки пересечения функции с осью абсцисс.в) ​возрастающей или убывающей является функция​

Answer 1

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Answer 2

1 а):f(-2)=4-16=-12
f(0)=0-16=-16

Answer 3

a) To find the values of f(-2) and f(0), simply substitute the given values of x into the function f(x) = x^2 - 16 and evaluate:

a.1) f(-2) = (-2)^2 - 16 f(-2) = 4 - 16 f(-2) = -12

a.2) f(0) = 0^2 - 16 f(0) = 0 - 16 f(0) = -16

b) To find the values of x for which f(x) equals -12 and 0, set the function equal to those values and solve for x:

b.1) f(x) = -12 x^2 - 16 = -12 x^2 = 4 x = ±√4 x = ±2

b.2) f(x) = 0 x^2 - 16 = 0 x^2 = 16 x = ±√16 x = ±4

So, the values of x for which f(x) equals -12 are x = 2 and x = -2, and the values of x for which f(x) equals 0 are x = 4 and x = -4.

For the function y = -2x + 1:

a) To find the coordinates of the point where the function intersects the x-axis, set y = 0 and solve for x:

0 = -2x + 1 2x = 1 x = 1/2

So, the point of intersection with the x-axis is (1/2, 0).

b) To determine if the function is increasing or decreasing, we need to examine the coefficient of x. In this case, the coefficient of x is -2. Since the coefficient is negative (-2), the function is decreasing. This means that as x increases, y (the value of the function) decreases.