help with Algebra 2 PROBLEM...PLEASE...?

Question

Without graphing, find the x- and y- intercepts of the graph of each equation.
*y = 2 √ 1-(x/4)^2
*x^2 + y^2 = 1
*y = √1-x^2
*y = - √ 1-x^2
*y = 3 √1-x^2
*y = √1-(2x)^2

Answer 1

My answers check with Kenyai's again, must work quicker than them in future lol

Answer 2

Knowing what the graph looks like helps.

y = 2√(1 - x²/16) is the top half of an ellipse, so there are 2 x intercepts when y=0 and 1 y intercept when x = 0. so,
0 = 2√(1 - x²/16)
0 = 1 - x²/16
x² = 16
x = ±4 ......... and
y = 2√(1 - 0/16)
y = 2.

#2 is a unit circle, x intercepts ±1, y intercepts ±1, again just letting x and y take turns being 0.

#3 is the top half of the unit circle, x = ±1, y = 1

#4 is bottom half of unit circle, x = ±1, y = -1

#5 again, top half of ellipse,
0 = 3√(1 - x²)
0 = 1 - x²
x = ±1,
and of course, when x = 0, y = 3.

#6 really, you should be able to look at these and do them in your head. if x = 0, y = ?, and if y = 0, x = ?

Answer 3

To find y intercepts, substitute 0 in for x.

To find x intercepts, substitute 0 in for y.

Y INTERCEPTS
1. (0, 2)
2. (0, 1) and (0, -1)
3. (0, 1)
4. (0, -1)
5. (0, 3)
6. (0, 1)

X INTERCEPTS
1. (4, 0) and (-4, 0)
2. (1, 0) and (-1, 0)
3. (1, 0) and (-1, 0)
4. (1, 0) and (-1, 0)
5. (1, 0) and (-1, 0)
6. (1/2, 0) and (-1/2, 0)

Answer 4

To find the x-intercepts, set y=0 and solve for x.
To find the y-intercepts, set x=0 and solve for y.

Answer 5

whoa.