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An equation for the line is x/(-2) + y/4 = 1 . I used the intercept-intercept form:
x/a + y/b = 1 where a and b denote the X and Y intercepts respectively. The corresponding function would be {(x,y)| x/(-2) + y/4 = 1; x and y any real nos.}
Y = x2
The function sec(x) is the secant function. It is related to the other functions by the expression 1/cos(x). It is not the inverse cosine or arccosine, it is one over the cosine function. Ex. cos(pi/4)= sqrt(2)/2 therefore secant is sec(pi/4)= 1/sqrt(2)/2 or 2/sqrt(2).
NO!!! It is an exponential function. 2 = 2^(1) 4 = 2^(2) 8 = 2^(3) 16 = 2^(4) 32 = 2^(5) ... 2^(n) If n = x on the co-ordinate axes, that is you go along 1,2,3,4,5 etc., the 'y' values (2,4,8,16,32, etc.,) , do NOT go up in a straight line, but an ever steepening gradient. Hence it is NOT a linear function.
{(-3, 2), (-4, 2), (8, 3), (7, 1)}
if you want to find the zero/null /x-intercept of any given function you have to find the x that makes it zero. therefore if f(x) = 2x +4 you have to put f(x) = 0 0 = 2x +4 2x = -4 x=-2 if you subsitute -2 for x you will find that the function f(x) will equal zero f(x)= 2 * (-2) +4 f(x) = -4 +4 f(x) = 0 if you have more complex functions like squared your approach will look something like this say g(x) is a function of x² g(x)=(x-2)² +4 set g(x)= 0 0= (x-2)² -4 4= (x-2) ² + or - radical (4) = x -2 2 + radical (4) = x or 2 -radical (4) = x x=4 or x = 0 remember radicals always have two solutions