Ask a New Question
Menu

11. What are the key points on the graph of y=x2−2x−120

? Name the vertex, x-intercepts, and y-intercept. Write your answers as ordered pairs. For the x-intercepts, write the leftmost point first. (4 points)
x-intercepts:
and

y-intercept:

vertex:

Question 2
12. A quadratic function is shown. f(x)=3x2−12x+11
. Complete the statements.(4 points)
Put responses in the correct input to answer the question. Select a response, navigate to the desired input and insert the response. Responses can be selected and inserted using the space bar, enter key, left mouse button or touchpad. Responses can also be moved by dragging with a mouse.
The function f(x)=3x2−12x+11
has a vertex of Response area and Response area is the equation of the axis of symmetry. The y-intercept of the function is Response area and one of the zeros is Response area rounded to the nearest tenth.
Skip to navigation
(3,12) (2,1) (11,1) x=3 x=2 x=11 (0,-1) (3.5, 0) (0,2.6) (0,11) (1.4,0) (11,0)

1 answer

Let's solve each question step by step.

Question 1

The given quadratic function is:

\[ y = x^2 - 2x - 120. \]

  1. Finding the vertex: The vertex form of a quadratic function is given by \( y = a(x - h)^2 + k \), where \( (h, k) \) is the vertex. Alternatively, the vertex can be found using the formula for \( h = -\frac{b}{2a} \).

    Here, \( a = 1 \) and \( b = -2 \).

    \[ h = -\frac{-2}{2 \cdot 1} = \frac{2}{2} = 1. \]

    To find \( k \), we substitute \( h \) back into the equation:

    \[ k = (1)^2 - 2(1) - 120 = 1 - 2 - 120 = -121. \]

    Therefore, the vertex is:

    Vertex: \( (1, -121) \)

  2. Finding the x-intercepts: To find the x-intercepts, we set \( y = 0 \):

    \[ 0 = x^2 - 2x - 120. \]

    We can factor this quadratic:

    \[ 0 = (x - 12)(x + 10). \]

    Setting each factor to zero gives:

    \[ x - 12 = 0 \Rightarrow x = 12, \] \[ x + 10 = 0 \Rightarrow x = -10. \]

    Thus, the x-intercepts are:

    x-intercepts: \( (-10, 0) \) and \( (12, 0) \) (where we list the leftmost point first).

  3. Finding the y-intercept: To find the y-intercept, we set \( x = 0 \):

    \[ y = (0)^2 - 2(0) - 120 = -120. \]

    Thus, the y-intercept is:

    y-intercept: \( (0, -120) \)

Summary for Question 1:

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

Question 2

The quadratic function is:

\[ f(x) = 3x^2 - 12x + 11. \]

  1. Finding the vertex: Using \( h = -\frac{b}{2a} \):

    Here, \( a = 3 \) and \( b = -12 \).

    \[ h = -\frac{-12}{2 \cdot 3} = \frac{12}{6} = 2. \]

    To find \( k \):

    \[ k = 3(2)^2 - 12(2) + 11 = 3(4) - 24 + 11 = 12 - 24 + 11 = -1. \]

    Thus, the vertex is:

    Vertex: \( (2, -1) \)

  2. Finding the axis of symmetry: The axis of symmetry is the vertical line that goes through the vertex, given by x = h.

    Axis of symmetry: \( x = 2 \)

  3. Finding the y-intercept: Setting \( x = 0 \):

    \[ f(0) = 3(0)^2 - 12(0) + 11 = 11. \]

    Thus, the y-intercept is:

    y-intercept: \( (0, 11) \)

  4. Finding the zeros (x-intercepts): We can apply the quadratic formula or factor:

    The discriminant \( b^2 - 4ac = (-12)^2 - 4(3)(11) = 144 - 132 = 12 \) (which is positive, indicating two real and distinct roots).

    Using the quadratic formula:

    \[ x = \frac{-(-12) \pm \sqrt{12}}{2 \cdot 3} = \frac{12 \pm 2\sqrt{3}}{6} = 2 \pm \frac{\sqrt{3}}{3}. \]

    Approximate the zeros, rounding to the nearest tenth:

    One zero is approximately \( 3.3 \) (using \( 2 + \frac{\sqrt{3}}{3} \)) and the other is approximately \( 1.7 \) (using \( 2 - \frac{\sqrt{3}}{3} \)).

Thus,

  • Vertex: \( (2, -1) \)
  • Axis of symmetry: \( x = 2 \)
  • Y-intercept: \( (0, 11) \)
  • One of the zeros: \( (1.7, 0) \) rounded to the nearest tenth.

Summary for Question 2:

  • Vertex: \( (2, -1) \)
  • Axis of symmetry: \( x = 2 \)
  • Y-intercept: \( (0, 11) \)
  • One of the zeros: \( (1.7, 0) \)