Choose ONE of the following equations:

x2−16x=−28

x2−16x−13=4

2(−x−2)=14x2

Then, in the space below:

SOLVE the equation showing all your work WITHOUT graphing.
Identify the following:
Vertex
X intercepts
Y intercepts
Show all your work for full credit. You can choose to do your work on paper and submit via webmail. If you submit via webmail write "Submitted via webmail" in the box below with the date.

You can choose to solve an additional equation for extra credit.

(5 points)

3 answers

Let's choose the equation \( x^2 - 16x = -28 \) to solve.

Step 1: Rewrite the equation

Rearranging the equation gives: \[ x^2 - 16x + 28 = 0 \]

Step 2: Use the Quadratic Formula

The quadratic formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -16 \), and \( c = 28 \).

Step 3: Calculate the discriminant

First, we calculate the discriminant \( b^2 - 4ac \): \[ b^2 - 4ac = (-16)^2 - 4(1)(28) = 256 - 112 = 144 \]

Step 4: Solve for x

Now we can substitute our values into the quadratic formula: \[ x = \frac{-(-16) \pm \sqrt{144}}{2(1)} = \frac{16 \pm 12}{2} \] Calculating the two potential values for \( x \):

  1. \( x = \frac{16 + 12}{2} = \frac{28}{2} = 14 \)
  2. \( x = \frac{16 - 12}{2} = \frac{4}{2} = 2 \)

Solutions

The solutions to the equation are: \[ x = 14 \quad \text{and} \quad x = 2 \]

Step 5: Find the Vertex

The vertex \( (h, k) \) of a parabola given by \( y = ax^2 + bx + c \) can be found using: \[ h = -\frac{b}{2a} = -\frac{-16}{2 \cdot 1} = \frac{16}{2} = 8 \] To find \( k \), substitute \( h = 8 \) back into the original equation: \[ k = 8^2 - 16(8) + 28 = 64 - 128 + 28 = -36 \] Thus, the vertex is: \[ (8, -36) \]

Step 6: Find the Y-intercept

To find the y-intercept, we set \( x = 0 \): \[ y = 0^2 - 16(0) + 28 = 28 \] The y-intercept is: \[ (0, 28) \]

Step 7: Summary of Results

  • Vertex: \( (8, -36) \)
  • X intercepts: \( (14, 0) \) and \( (2, 0) \)
  • Y intercept: \( (0, 28) \)

This concludes the solution to the selected equation. If this were to be submitted via webmail, I would note "Submitted via webmail" and include the date of submission.

all in one simple sentence

The solutions to the equation \( x^2 - 16x = -28 \) are \( x = 14 \) and \( x = 2 \), with the vertex at \( (8, -36) \), x-intercepts at \( (14, 0) \) and \( (2, 0) \), and a y-intercept at \( (0, 28) \).