Let's evaluate each student's claim regarding the function \( f(x) = (x + 3)(x + 5) \).
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Jeremiah: The y-intercept is at (15, 0).
- To find the y-intercept, we set \( x = 0 \): \[ f(0) = (0 + 3)(0 + 5) = 3 \times 5 = 15. \]
- The y-intercept is at (0, 15), not (15, 0). Jeremiah's claim is incorrect.
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Lindsay: The x-intercepts are at (–3, 0) and (5, 0).
- To find the x-intercepts, we set \( f(x) = 0 \): \[ (x + 3)(x + 5) = 0. \]
- This gives us the solutions: \[ x + 3 = 0 \implies x = -3, \] \[ x + 5 = 0 \implies x = -5. \]
- Therefore, the x-intercepts are (–3, 0) and (–5, 0). Lindsay's claim is incorrect.
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Stephen: The vertex is at (–4, –1).
- To find the vertex of a quadratic function in the form \( ax^2 + bx + c \), we can convert \( f(x) \) to standard form or use the vertex formula \( x = -\frac{b}{2a} \).
- First, let's expand \( f(x) \): \[ f(x) = x^2 + 8x + 15. \]
- Here, \( a = 1 \), \( b = 8 \), and \( c = 15 \). The x-coordinate of the vertex is: \[ x = -\frac{8}{2 \times 1} = -4. \]
- Finding the y-coordinate of the vertex: \[ f(-4) = (-4 + 3)(-4 + 5) = (-1)(1) = -1. \]
- Therefore, the vertex is at (–4, –1). Stephen's claim is correct.
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Alexis: The midpoint between the x-intercepts is at (4, 0).
- The x-intercepts are \( x = -3 \) and \( x = -5 \).
- The midpoint \( m \) of two points \( x_1 \) and \( x_2 \) is given by: \[ m = \frac{x_1 + x_2}{2} = \frac{-3 + (-5)}{2} = \frac{-8}{2} = -4. \]
- Thus, the midpoint of the x-intercepts is at (–4, 0), not (4, 0). Alexis's claim is incorrect.
Therefore, the claim by Stephen is correct.
1 answer