To answer you:
The graph has x-intercepts at x = 6
and x = -2, a y-intercept at y = -3/2,
and a horizontal asymptote at y = 0.
The graph opens upward and is
symmetric about the vertical line x = 2.
EXPLANATION:
To graph the equation y = 1/8 (X-6)
(x+2), we can start by finding the x-and y-intercepts, as well as any vertical or horizontal asymptotes.
To find the x-intercepts, we set y = 0
and solve for x:
0 = 1/8(x-6)(×+2)
0 = (x-6)(x+2)
So the x-intercepts are x = 6 and x =
-2.
To find the y-intercept, we set x = 0
and solve for y:
y = 1/8(0-6) (0+2)
y = 1/8(-6) (2)
y= -3/2
So the y-intercept is y = -3/2.
To determine any vertical or horizontal asymptotes, we can look at the degree of the numerator and denominator of the rational function. In this case, the numerator has degree 2 and the denominator has degree 0, so there are no vertical asymptotes and the
horizontal asymptote is y = 0.
With this information, we can sketch a graph of the equation:
The graph has x-intercepts at x = 6
and x = -2, a y-intercept at y = -3/2,
and a horizontal asymptote at y = 0.
The graph opens upward and is
symmetric about the vertical line x = 2.