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2. 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! Below are the detailed steps to solve each problem.

Question 2: Complete the square to determine the vertex and axis of symmetry of \( y = -x^2 + 8x + 7 \)

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

  2. Factor out the coefficient of \( x^2 \): Since the coefficient of \( x^2 \) is negative, factor out -1 from the first two terms: \[ y = - (x^2 - 8x) + 7 \]

  3. Complete the square: Take the coefficient of \( x \) (which is -8), halve it to get -4, and square it to get 16. Add and subtract this value inside the parentheses: \[ y = - (x^2 - 8x + 16 - 16) + 7 \] \[ y = - ((x - 4)^2 - 16) + 7 \] \[ y = - (x - 4)^2 + 16 + 7 \] \[ y = - (x - 4)^2 + 23 \]

  4. Identify the vertex: The vertex form is \( y = a(x - h)^2 + k \). Here, \( h = 4 \) and \( k = 23 \), so the vertex is \( (4, 23) \).

  5. Determine the axis of symmetry: The axis of symmetry is the vertical line that goes through the vertex. It is given by: \[ x = h \] So the axis of symmetry is \( x = 4 \).

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

  1. Quadratic formula: The zeros of the quadratic equation \( ax^2 + bx + c = 0 \) can be found using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 6 \), \( b = 13 \), and \( c = 5 \).

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

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

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

    1. For the plus sign: \[ x = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2} \]
    2. For the minus sign: \[ x = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3} \]

  5. Conclusion - zeros are: \[ x = -\frac{1}{2} \quad \text{and} \quad x = -\frac{5}{3} \]

Summary of Results:

  • Problem 2: The vertex is \( (4, 23) \) and the axis of symmetry is \( x = 4 \).
  • Problem 3: The zeros of the quadratic equation are \( x = -\frac{1}{2} \) and \( x = -\frac{5}{3} \).

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