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) x -intercept: left parenthesis 0 comma negative 120 right parenthesis y -intercept: left parenthesis negative 10 comma 0 right parenthesis left parenthesis 12 comma 0 right parenthesis vertex: left parenthesis 1 comma negative 121 right parenthesis Skip to navigation
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)
To find the key points on the graph of the equation y = x^2 - 2x - 120, we can start by determining the x-intercepts. These are the points where the graph intersects the x-axis.
To find the x-intercepts, we set y = 0 in the equation and solve for x:
0 = x^2 - 2x - 120
Now, we can factorize the quadratic equation:
0 = (x - 12)(x + 10)
Setting each factor equal to zero, we find two x-intercepts:
x - 12 = 0 --> x = 12
x + 10 = 0 --> x = -10
So, the x-intercepts are (-10, 0) and (12, 0).
Next, let's find the y-intercept. This is the point where the graph intersects the y-axis.
To find the y-intercept, we set x = 0 in the equation and solve for y:
y = (0)^2 - 2(0) - 120
y = -120
Therefore, the y-intercept is (0, -120).
Lastly, let's find the vertex. The vertex of a quadratic equation in the form y = ax^2 + bx + c can be found using the formula x = -b/2a.
In this case, a = 1 and b = -2. Plugging these values into the formula:
x = -(-2) / (2 * 1)
x = 2 / 2
x = 1
To find the y-coordinate of the vertex, we substitute the x-value into the equation:
y = (1)^2 - 2(1) - 120
y = 1 - 2 - 120
y = -121
Therefore, the vertex is (1, -121).
In summary, the key points on the graph of the equation y = x^2 - 2x - 120 are:
- x-intercepts: (-10, 0) and (12, 0)
- y-intercept: (0, -120)
- vertex: (1, -121)