Fog Zero
help me with these precalculus problems?
Find the equation of the line that passes through (1,3) and is perpendicular to the line 2x+3y+5=0
Find the range of the function f(x)= SqRoot x^2-9
Given f(x)= 1/x^2 and g(x) = SqRoot x^2+4, find (FoG)(x)
Find all the real zeroes of the polynomial function f(x) =x^3-3x^2-4x
Find the vertical asymptote(s) f(x)= x+2/x^2-9
1) perpendicular means slope of one line = -1/slope of second line
first line, (the one given)
2x+3y+5=0
3y = -2x-5
y=-2/3 x -5/3
slope = -2/3
2nd line,
y = mx + b
where m = -1/(2/3) = 3/2
so y = 3/2 x + b
plug in your x and y values and find b
3 = 3/2 *(1) + b
b = 3 - 3/2 = 3/2
so y = 3/2 x + 3/2
in general form, 3x -2y + 3 = 0
(2) range of function...
f(x) = (x^2 -9)^.5
notice the square root? this function is valid only when
(x^2-9) >= 0
so x^2 >= 9
therefore x >=3, and x < = -3
next find f(x) for x>-3 and x< =-3
notice f(x) >=0 for all cases..
ie range is f(x) >=0
(3) f(g(x)).... (you called it FoG. same thing)
plug and chug...
f(g(x)) = 1/(g(x))^2 = 1/((x^2+4)^.5)^2 = 1/(x^2+4)
(4) find all zeros.....
f(x) = x^3 - 3x^2 -4x.
set f(x) = 0 and factor...
0 = (x) * ( x^2 -3x -4) = (x) * (x -4) * (x+1)
so zeros are when x = 0, when x-4 = zero, and when x+1 = 0
ie x = 0, 4, and -1
double check. plug in 0,4,-1 in the function and calc...
(5) vertical asymptotes
occur when denominator of function = 0 (since any number / 0 = undetermined...)
I'm going to assume your f(x) is this.... if not, then you correct
f(x) = (x+2)/(x^2-9)
asymptotes occur when
x^2 - 9 = 0
x^2 = 9
x = 3 and x = -3