y = -3/2x + 1
y = (-3 + 2x)/2x or
y = (2x - 3)/2x
We have to deal with a rational function here f(x) = p(x)/q(x), where 2x ≠ 0, or x ≠ 0. So that the domain of f is the st of all real numbers except 0, and the y-axis (the line x = 0) is a vertical asymptote of the graph of f.
Since the degree of the numerator and denominator is the same (degree 1), then the line y = 2/2 (an/bn) or y = 1, is the horizontal asymptote of the graph of f.
y = (2x - 3)/2x
x = 0, gives us the y-intercept, but in our case x cannot be zero, so we don't have y-intercept here.
y = 0, gives us the x-intercept which is 3/2.
y = (2x - 3)/2x
0 = (2x - 3)/2x
0 = 2x - 3
0 + 3 = 2x - 3 + 3
3 = 2x
3/2 = 2x/2
3/2 = x
Since f(-x) ≠ f(x) or -f(x), then the the graph of f has nether y-axis nor origin symmetry.
y = -3/2x + 1 or
f(x) = -3/2x + 1
f(-x) = -3/2(-x) + 1
f(-x) = 3/2x + 1
-f(x) = -(-3/2x + 1)
-f(x) = 3/2x - 1
We can graph the function by plotting points between and beyond the x-intercept (x = 3/2) and vertical asymptote (x = 0).
So we can valuate the function at -2, -1, 1, and 2.
Graph the function:
1
2
(1, 9)
(3, 1)
y=x+1 there for answer is 2
x = 1 and y = -2
y=10
They are expressions.
1
Y = 2x + 1
It is: 1
The solution is: x = 1 and y = -1