What are the key points on the graph of y=x^2−2x−120

Question

What are the key points on the graph of  y=x^2−2x−120
 ? Name the vertex, x-intercept(s), and y-intercept.(1 point)
Responses

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

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

GPT-4o mini · Answer

To find the key points of the quadratic function $y = x^{2} - 2 x - 120$ , we can go through the process of identifying the vertex, x-intercepts, and y-intercept.

Vertex: The vertex can be found using the formula $x = - \frac{b}{2 a}$ for the quadratic equation $a x^{2} + b x + c$ . Here, $a = 1$ and $b = - 2$ .

$x = - \frac{- 2}{2 \cdot 1} = \frac{2}{2} = 1$

Now substituting $x = 1$ back into the equation to find the y-coordinate:

$y = (1)^{2} - 2 (1) - 120 = 1 - 2 - 120 = - 121$

So the vertex is $(1, - 121)$ .
X-Intercepts: To find the x-intercepts, set $y = 0$ :

$0 = x^{2} - 2 x - 120$

Factoring the quadratic:

$(x - 12) (x + 10) = 0$

Therefore, the solutions are:

$x = 12 and x = - 10$

So the x-intercepts are $(12, 0)$ and $(- 10, 0)$ .
Y-Intercept: To find the y-intercept, set $x = 0$ :

$y = (0)^{2} - 2 (0) - 120 = - 120$

So the y-intercept is $(0, - 120)$ .

Summary of Key Points:

Vertex: $(1, - 121)$
X-Intercepts: $(- 10, 0)$ and $(12, 0)$
Y-Intercept: $(0, - 120)$

Final response:

x-intercepts: $(- 10, 0), (12, 0)$
y-intercept: $(0, - 120)$
vertex: $(1, - 121)$