Find the coordinates of the x- and y-intercepts for an ellipse with the equation (x+1)^2/9 + (y-2)^2/8 = 1
1 answer:
Answer:
- x-intercepts: -1±√4.5
- y-intercepts: 2±8/3
Step-by-step explanation:
The x-intercepts are where y = 0.
(x +1)^2/9 +(0 -2)^2/8 = 1
(x +1)^2/9 +1/2 = 1 . . . . . simplify a bit
(x +1)^2 = 9/2 . . . . . . . . .subtract 1/2, multiply by 9; next: square root, add -1
x = -1 ±√4.5 . . . . . x-intercepts
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The y-intercepts are where x=0.
(0+1)^2/9 +(y-2)^2/8 = 1
1/9 + (y -2)^2/8 = 1 . . . . . simplify a bit
(y -2)^2 = 64/9 . . . . . . . . subtract 1/9, multiply by 8,
y = 2 ±8/3 . . . . . . take the square root, add 2 . . . . y-intercepts
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Answer:
Step-by-step explanation:
First create an equation in slope-intercept form, y = mx + b. To do this, find the rate of change or slope (m), and the y-intercept (b).
Using two pairs, (-7, -3) and (-8, -4),
m = 1
To find b, substitute m = 1, x = -7 and y = -3 into y = mx + b.
Thus:
-3 = (1)(-7) + b
-3 = -7 + b
-3 + 7 = b
4 = b
b = 4.
Replace y with f(x), m with 1 and b with 4 in y = mx + b.
The function would be: