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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:

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15y ago
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Q: Y equals -3over2x plus 1
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