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Question 1 A)Determine the zeros of the quadratic function y=x2+15x+36y=x2+15x+36. (When listing the smaller zero, remember that negative values are always smaller than positive values.)(1 point) smaller zero: x=x=$$; greater zero: x=x=$$ Question 2 A)Determine the zeros of the quadratic function y=−x2+16x−39y=−x2+16x−39. (When listing the smaller zero, remember that negative values are always smaller than positive values.)(1 point) smaller zero: x=x=$$; greater zero: x=x=$$ Question 3 A)Determine the vertex and axis of symmetry of y=x2−10x+3y=x2−10x+3.(1 point) vertex: ($$); axis of symmetry: x=x=$$ Question 4 A)What point is symmetrical to (−1,−3)(−1,−3) for the graph of y=x2+12x+8y=x2+12x+8?(1 point) ($$) Question 5 A)What are the key points on the graph of y=x2−16x+48y=x2−16x+48? Name the vertex, x-intercepts, and y-intercept.(2 points) vertex: ($$) (smaller) x-intercept: ($$) (larger) x-intercept: ($$) y-intercept: ($$) page 19 of 20

1 answer

Let's solve each of the questions step by step.

Question 1

We need to find the zeros of the quadratic function \( y = x^2 + 15x + 36 \).

To find the zeros, we can factor the quadratic: \[ x^2 + 15x + 36 = (x + 3)(x + 12) = 0 \] Setting each factor to zero gives: \[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \] \[ x + 12 = 0 \quad \Rightarrow \quad x = -12 \] Thus, the zeros are:

  • smaller zero: \( x = -12 \)
  • greater zero: \( x = -3 \)

Question 2

Now we find the zeros of the quadratic function \( y = -x^2 + 16x - 39 \).

We can set the equation to zero: \[ -x^2 + 16x - 39 = 0 \quad \text{(multiply by -1 for simplicity)} \] \[ x^2 - 16x + 39 = 0 \] Next, we can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1, b = -16, c = 39 \): \[ x = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot 39}}{2 \cdot 1} = \frac{16 \pm \sqrt{256 - 156}}{2} = \frac{16 \pm \sqrt{100}}{2} = \frac{16 \pm 10}{2} \]

Calculating the solutions: \[ x = \frac{26}{2} = 13 \quad \text{(greater zero)} \] \[ x = \frac{6}{2} = 3 \quad \text{(smaller zero)} \] Thus, the zeros are:

  • smaller zero: \( x = 3 \)
  • greater zero: \( x = 13 \)

Question 3

For the vertex and axis of symmetry of \( y = x^2 - 10x + 3 \):

The x-coordinate of the vertex is given by: \[ x = -\frac{b}{2a} = -\frac{-10}{2 \cdot 1} = \frac{10}{2} = 5 \] To find the y-coordinate of the vertex, substitute \( x = 5 \) into the equation: \[ y = (5)^2 - 10(5) + 3 = 25 - 50 + 3 = -22 \] Thus, the vertex is \( (5, -22) \) and the axis of symmetry is \( x = 5 \).

Question 4

We need to find the symmetrical point to \( (-1, -3) \) for the graph of \( y=x^2 + 12x + 8 \).

First, we calculate the vertex of the parabola: \[ x = -\frac{12}{2} = -6 \] Next, we find the y-coordinate of the vertex: \[ y = (-6)^2 + 12(-6) + 8 = 36 - 72 + 8 = -28 \] The vertex is \( (-6, -28) \). The line of symmetry is \( x = -6 \).

To find the point symmetrical to \( (-1, -3) \), we compute the distance from \( -1 \) to \( -6 \): \[ -6 - (-1) = -5 \] We move the same distance to the left of \( -6 \): \[ -6 - 5 = -11 \] So the symmetrical point is \( (-11, -3) \).

Question 5

We find the key points for the graph of \( y = x^2 - 16x + 48 \).

  1. Vertex: The x-coordinate is: \[ x = \frac{16}{2} = 8 \] Now substitute \( x = 8 \) into the function: \[ y = (8)^2 - 16(8) + 48 = 64 - 128 + 48 = -16 \] So, the vertex is \( (8, -16) \).

  2. X-intercepts: Set \( y = 0 \): \[ x^2 - 16x + 48 = 0 \] Factoring: \[ (x - 12)(x - 4) = 0 \] This gives \( x = 12 \) and \( x = 4 \).

  3. Y-intercept: Set \( x = 0 \): \[ y = 0^2 - 16(0) + 48 = 48 \] Thus the y-intercept is \( (0, 48) \).

Final Answers

  • Question 1: smaller zero: \( x = -12 \); greater zero: \( x = -3 \)
  • Question 2: smaller zero: \( x = 3 \); greater zero: \( x = 13 \)
  • Question 3: vertex: \( (5, -22) \); axis of symmetry: \( x = 5 \)
  • Question 4: point: \( (-11, -3) \)
  • Question 5: vertex: \( (8, -16) \); smaller x-intercept: \( (4, 0) \); larger x-intercept: \( (12, 0) \); y-intercept: \( (0, 48) \)