find any intercepts of the graph of the given equation. determine whether the graph of the equation possesses symmetry with respect to the x-axis, y-axis, or orgin
y=x^2-2x-2
Already answered
To find the intercepts of the graph of the equation y = x^2 - 2x - 2, we set y to zero and solve for x.
1. Y-intercept (where the graph intersects the y-axis): Set x = 0.
y = (0)^2 - 2(0) - 2
y = -2
Therefore, the y-intercept is (0, -2).
2. X-intercepts (where the graph intersects the x-axis): Set y = 0.
0 = x^2 - 2x - 2
To solve this quadratic equation, you can either factor it or use the quadratic formula.
Let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = 1, b = -2, and c = -2.
x = (-(-2) ± √((-2)^2 - 4(1)(-2))) / (2(1))
x = (2 ± √(4 + 8)) / 2
x = (2 ± √12) / 2
x = (2 ± 2√3) / 2
x = 1 ± √3
Therefore, the x-intercepts are (1 + √3, 0) and (1 - √3, 0).
To determine whether the graph of the equation possesses symmetry with respect to the x-axis, y-axis, or origin:
1. X-axis symmetry: If replacing y with -y in the equation results in an equivalent equation, then the graph possesses x-axis symmetry.
Let's try it: -y = x^2 - 2x - 2
Multiply through by -1: y = -x^2 + 2x + 2
The equation is not equivalent, so the graph does not possess x-axis symmetry.
2. Y-axis symmetry: If replacing x with -x in the equation results in an equivalent equation, then the graph possesses y-axis symmetry.
Let's try it: y = (-x)^2 - 2(-x) - 2
Simplify: y = x^2 + 2x - 2
The equation is not equivalent, so the graph does not possess y-axis symmetry.
3. Origin symmetry: If replacing x with -x and y with -y in the equation results in an equivalent equation, then the graph possesses origin symmetry.
Let's try it: -y = (-x)^2 - 2(-x) - 2
Multiply through by -1: y = x^2 - 2x - 2
The equation is not equivalent, so the graph does not possess origin symmetry.
Therefore, the graph of the equation y = x^2 - 2x - 2 has no symmetry with respect to the x-axis, y-axis, or origin.