X and Y intercepts??

Question

Im working on a review sheet for AP Calculus and cannot remember how to find X and Y-Intercepts.
(x^2/9) + (y^2/16) = 1
Please explain how to solve this

Answer 1

X intercept is when the graph touches the X-axis (so Y is zero) and vice versa.

So for X-intercept, put y=0 in the equation.
=> x^2/9 + 0 = 1
=> x^2/9 = 1
=> x^2 = 9
=> x = Square root of 9
=> x = 3 or -3
So X-intercepts are (3,0) & (-3,0), i.e., the graph touches X-axis at the above two points.

Similarly for Y-intercept, put x = 0.
=> y^2/16 + 0 = 1
=> y^2/16 = 1
=> y^2 = 16
=> y = Square root of 16
=> y = 4 or -4
So Y-intercepts are (0,4) & (0,-4), i.e., the graph touches Y-axis at the above two points.

Answer 2

The x-intercepts of a curve are all the places where the curve crosses ("intercepts") the x-axis -- and since every point on the x-axis has a y-coordinate of 0, to find the x-intercepts, you substitute 0 for y and solve what's left for x.

Similarly, to find the y-intercepts, you substitute 0 for x and solve what's left for y.

For example, in your case, x²/9 + y²/16 = 1, if we plug in 0 for y, we're left with x²/9 = 1, which can be easily solved for x, giving us the x-intercept values of 3 and -3 -- or, more accurately, the points (3,0) and (-3,0).

Hope that gives you the idea. Good luck!

Answer 3

(x^2/9) + (y^2/16) = 1
This is the equation of ellipse
In order to find the x- intercept, put y=0 because on the x -axis y=0
(x^2/9) + (0/16) = 1
(x^2/9) = 1
x^2 = 9
Therefore x=-3 and x =3
Hence the X- intercepts are (3,0) and (-3,0)
Similarly for finding Y- intercept put x=0, and solve
you will find y-intercept (0,4) and (0,-4)

Answer 4

You wrote: 2x--10=0 it really is inaccurate. 2x - 3(0) -10=0 2x - 10 = 0 2x = 10 x = 5 So the x intercept is 5. to discover the y-intercept, set x = 0 2(0) - 3y -10 = 0 -3y-10=0 -10 = 3y y = -10/3 So the y intercept is - 10/3 or -3 a million/3

Answer 5

X intercepts=+/-3 and Y intercepts +/-4

Answer 6

THe simplest way is to set x equals 0, to find the y intercept. And vice versa for the x intercept.

Answer 7

solve for y = 0 and x = 0.

Answer 8

try this:
multiplicate the ecuation for 144:
result
32x+18y=144 and then
18y=144-32x
y=(114/18)-(32x/18)
like this you will find the intercepts in y and then chage for x to do it.

Answer 9

((x^2)/9) + ((y^2)/16) = 1

((0^2)/9) + ((y^2)/16) = 1

((y^2)/16) = 1
y^2 = 16
y = -4 or 4

((x^2)/9) + ((0^2)/16) = 1
((x^2)/9) = 1
x^2 = 9
x = 3 or -3

ANS : (0,-4), (0,4), (3,0), and (-3,0)

Answer 10

the X- intercepts are (3,0) and (-3,0)
the Y-intercepts are (0,4) and (0,-4)