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 \( y = -x^2 + 8x + 7 \)
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Start with the function: \[ y = -x^2 + 8x + 7 \]
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Factor out the coefficient of \( x^2 \) (which is -1): \[ y = -1(x^2 - 8x) + 7 \]
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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.
\[ y = -1(x^2 - 8x + 16 - 16) + 7 \] \[ y = -1((x - 4)^2 - 16) + 7 \]
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Simplify: \[ y = -1(x - 4)^2 + 16 + 7 \] \[ y = -1(x - 4)^2 + 23 \]
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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 \( (4, 23) \).
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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 \( y = 6x^2 + 13x + 5 \)
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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 \( c = 5 \).
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Calculate the discriminant: \[ b^2 - 4ac = 13^2 - 4 \cdot 6 \cdot 5 \] \[ = 169 - 120 = 49 \]
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Substituting into the formula: \[ x = \frac{-13 \pm \sqrt{49}}{2 \cdot 6} \] \[ x = \frac{-13 \pm 7}{12} \]
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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!