Complete the square to determine the vertex and axis of symmetry of y=−x2+8x+7

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3. Find the zeros of the quadratic equation y=6x2+13x+5
. You may use any of the following methods: factoring, square roots, complete the square, or quadratic formula.

Please show all steps to solve these two questions. Upload a picture of your work for each and clearly label them as #2 and #3. You may use Desmos to check your work, but you must upload your handwritten work to receive credit.

1 answer

Sure! Let’s complete the square for the first equation and find the vertex and axis of symmetry. Then we’ll find the zeros of the quadratic equation in the second part.

Problem 2: Complete the square for \( y = -x^2 + 8x + 7 \)

  1. Start with the function: \[ y = -x^2 + 8x + 7 \]

  2. Factor out the coefficient of \( x^2 \) (which is -1): \[ y = -1(x^2 - 8x) + 7 \]

  3. Complete the square:

    • Take half of the coefficient of \( x \) (which is -8), square it, and add/subtract inside the parentheses:
    • Half of -8 is -4, and squaring it gives us 16.

    \[ y = -1(x^2 - 8x + 16 - 16) + 7 \] \[ y = -1((x - 4)^2 - 16) + 7 \]

  4. Simplify: \[ y = -1(x - 4)^2 + 16 + 7 \] \[ y = -1(x - 4)^2 + 23 \]

  5. Identify the vertex: The vertex form of the equation is \( y = a(x - h)^2 + k \), where \( (h, k) \) is the vertex.

    • Here, \( h = 4 \) and \( k = 23 \), so the vertex is \( (4, 23) \).
  6. Determine the axis of symmetry: The axis of symmetry can be found using \( x = h \).

    • So, the axis of symmetry is \( x = 4 \).

Answer for Problem 2:

  • Vertex: \( (4, 23) \)
  • Axis of symmetry: \( x = 4 \)

Problem 3: Find the zeros of \( y = 6x^2 + 13x + 5 \)

  1. Use the quadratic formula: The quadratic formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 6 \), \( b = 13 \), and \( c = 5 \).

  2. Calculate the discriminant: \[ b^2 - 4ac = 13^2 - 4 \cdot 6 \cdot 5 \] \[ = 169 - 120 = 49 \]

  3. Substituting into the formula: \[ x = \frac{-13 \pm \sqrt{49}}{2 \cdot 6} \] \[ x = \frac{-13 \pm 7}{12} \]

  4. Calculate the two possible values for \( x \):

    • For \( x_1 \): \[ x_1 = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2} \]
    • For \( x_2 \): \[ x_2 = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3} \]

Answer for Problem 3:

  • Zeros: \( x = -\frac{1}{2}, x = -\frac{5}{3} \)

The final step is to write these down on paper, show the calculations clearly, and upload images labeled as #2 and #3, as requested. If you need assistance formatting that or have further questions, feel free to ask!