The horizontal asymptote at y=1 indicates that the numerator and denominator are of the same degree.
The vertical asymptotes indicate that
y = p(x) / (x+5)(x-4)
for some quadratic polynomial p(x)
The x-intercepts then dictate that
y = a(x+6)(x-5) / (x+5)(x-4)
The y-intercept at (0,3/2) means that
a(6)(-5) / (5)(-4) = 3/2
a = 3/2 * 20/30 = 1
So, y = (x+6)(x-5) / (x+5)(x-4) = (x^2+x-30)/(x^2+x-20)
Find a possible formula for the function graphed below. The x-intercepts are marked with points located at (5, 0) and (-6, 0), while the y-intercept is marked with a point located at \left( 0, \frac{3}{2} \right). The asymptotes are y = 1, x = -5, and x = 4. Give your formula as a reduced rational function.
2 answers
x-intercepts are at (5, 0) and (-6, 0)
y = (x-5)(x+6)
asymptotes are x = -5, and x = 4.
y = (x-5)(x+6) / (x+5)(x-4)
asymptote at y = 1
still good for that, since top and bottom are the same degree, with highest coefficients both = 1
y-intercept is (0,3/2)
so let's check what y is when x=0.
y = (-5)(6) / (5)(-4) = -30/-20 = 3/2
so our final function is
y = (x-5)(x+6) / (x+5)(x-4)
so -- extra credit. How would you change the y-intercept, without changing any of the other parts?
y = (x-5)(x+6)
asymptotes are x = -5, and x = 4.
y = (x-5)(x+6) / (x+5)(x-4)
asymptote at y = 1
still good for that, since top and bottom are the same degree, with highest coefficients both = 1
y-intercept is (0,3/2)
so let's check what y is when x=0.
y = (-5)(6) / (5)(-4) = -30/-20 = 3/2
so our final function is
y = (x-5)(x+6) / (x+5)(x-4)
so -- extra credit. How would you change the y-intercept, without changing any of the other parts?