What are the key points on the graph of y=x2−2x−120
? Name the vertex, x-intercept(s), and y-intercept.(1 point)
Responses
x-intercepts: (−10,0)
(12,0)
y-intercept: (1,−121)
vertex: (0,−120)
x -intercepts: left parenthesis negative 10 comma 0 right parenthesis left parenthesis 12 comma 0 right parenthesis y-intercept: left parenthesis 1 comma negative 121 right parenthesis vertex: left parenthesis 0 comma negative 120 right parenthesis
x-intercepts: (−10,0)
(12,0)
y-intercept: (0,−120)
vertex: (1,−121)
x -intercepts: left parenthesis negative 10 comma 0 right parenthesis left parenthesis 12 comma 0 right parenthesis y -intercept: left parenthesis 0 comma negative 120 right parenthesis vertex: left parenthesis 1 comma negative 121 right parenthesis
x-intercept: (1,−121)
y-intercept: (0,−120)
vertex: (−10,0)
(12,0)
x -intercept: left parenthesis 1 comma negative 121 right parenthesis y -intercept: left parenthesis 0 comma negative 120 right parenthesis vertex: left parenthesis negative 10 comma 0 right parenthesis left parenthesis 12 comma 0 right parenthesis
x-intercept: (0,−120)
y-intercept: (−10,0)
(12,0)
vertex: (1,−121)
The correct answer is:
x-intercepts: (-10,0) and (12,0)
y-intercept: (0,-120)
vertex: (1,-121)
The correct answer is:
x-intercepts: (-10, 0) and (12, 0)
y-intercept: (0, -120)
vertex: (1, -121)
The key points on the graph of the equation y=x^2-2x-120 are as follows:
x-intercepts: The x-intercepts are the points where the graph intersects the x-axis. To find the x-intercepts, we set y equal to zero and solve for x. In this case, we need to solve the equation x^2-2x-120=0. The solutions to this quadratic equation are x=-10 and x=12. Therefore, the x-intercepts are (-10,0) and (12,0).
y-intercept: The y-intercept is the point where the graph intersects the y-axis. To find the y-intercept, we set x equal to zero and solve for y. When x=0, we have y=0^2-2(0)-120=-120. Therefore, the y-intercept is (0,-120).
vertex: The vertex is the highest or lowest point on the graph, depending on the direction the parabola opens. To find the vertex, we can use the formula x = -b/2a, where a and b are the coefficients of the quadratic equation. In this case, a=1 and b=-2. Plugging these values into the formula, we get x = -(-2)/(2*1) = 1. To find the corresponding y-coordinate, we substitute x=1 into the equation y=x^2-2x-120. We get y=1^2-2(1)-120=-121. Therefore, the vertex is (1,-121).
To summarize, the key points on the graph of y=x^2-2x-120 are:
- x-intercepts: (-10,0) and (12,0)
- y-intercept: (0,-120)
- vertex: (1,-121)