No it does not intersect the y-axis https://www.desmos.com/calculator got to this site plug in that equation it will give you graph,you can see its not intersecting the y-axis
First, write the equation in standard form by completing the square.
x2 - 4x + y2 = -3
x2 - 4x + 4 + y2 = -3 + 4
(x - 2)2 + y2 = 1
The circle is centered at (2, 0) with a radius of 1. Since the circle is centered on the x-axis, it intersects the x-axis two times, at (3, 0) and (1, 0).
No. The y-intercept is the value of y when x=0, hence it intersects the y-axis when x=0. In this case:
0^2-4(0)+y^2=-3
y^2=-3
y=√-3
Since there is no real square root of a negative value, the curve does not intersect the y-axis
A
Step-by-step explanation:
The answer is C) No, because the circle is entirely in the first quadrant
No it a first quadrant
No it does not intersect the y-axis
https://www.desmos.com/calculator
got to this site plug in that equation it will give you graph,you can see its not intersecting the y-axis
Step-by-step explanation:
B) No, because the center is a (2, 0) and the radius is 1
Yes, because the center is on the x-axis.
Step-by-step explanation:
First, write the equation in standard form by completing the square.
x2 - 4x + y2 = -3
x2 - 4x + 4 + y2 = -3 + 4
(x - 2)2 + y2 = 1
The circle is centered at (2, 0) with a radius of 1. Since the circle is centered on the x-axis, it intersects the x-axis two times, at (3, 0) and (1, 0).
Yes, twice
Step-by-step explanation:
The equation will intersect the x-axis when y = 0, so we have
x² - 4x = -3 now solve this quadratic for x...
x² - 4x + 3 = 0
factor...
(x - 3)(x - 1) = 0,
so at x = 1 and x = 3, the function crosses the x-axis
See the graph below
[tex]20 ! does the equation x2 - 4x + y2 = -3 intersect the x-axis?[/tex]
Yes, because the center is on the x-axis, the circle will intersect 2 times, at x=1 and at x=3
No, because the center is a (2, 0) and the radius is 1.
Step-by-step explanation: