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Determine the quadratic function whose graph is given. 6) -5 (-1,-1) I (0, 1) 5x (0

Determine the quadratic function whose graph is given. 6) -5 (-1,-1) I (0, 1) 5x (0

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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**Determine the quadratic function whose graph is given.** The graph shows a parabola with a vertex at the point (-1, -1) and it also passes through the point (0, 1). The parabola opens upwards, indicating a positive leading coefficient. **Options for the quadratic function:** - **A)** \( f(x) = -2x^2 + 4x + 1 \) - **B)** \( f(x) = -2x^2 - 4x - 1 \) - **C)** \( f(x) = 2x^2 + 4x + 1 \) - **D)** \( f(x) = 2x^2 - 8x + 1 \) **Explanation of the graph:** - **Vertex:** The graph has a vertex at \((-1, -1)\). - **Y-intercept:** The graph crosses the y-axis at \((0, 1)\). To determine which function corresponds to the graph, consider the signs and values of the coefficients. Since the parabola opens upwards, the leading coefficient must be positive. Additionally, the vertex form or intercepts can be used to verify the correct function. Based on this, the function that matches the graph is likely **C**: \( f(x) = 2x^2 + 4x + 1 \).
Transcribed Image Text:**Determine the quadratic function whose graph is given.** The graph shows a parabola with a vertex at the point (-1, -1) and it also passes through the point (0, 1). The parabola opens upwards, indicating a positive leading coefficient. **Options for the quadratic function:** - **A)** \( f(x) = -2x^2 + 4x + 1 \) - **B)** \( f(x) = -2x^2 - 4x - 1 \) - **C)** \( f(x) = 2x^2 + 4x + 1 \) - **D)** \( f(x) = 2x^2 - 8x + 1 \) **Explanation of the graph:** - **Vertex:** The graph has a vertex at \((-1, -1)\). - **Y-intercept:** The graph crosses the y-axis at \((0, 1)\). To determine which function corresponds to the graph, consider the signs and values of the coefficients. Since the parabola opens upwards, the leading coefficient must be positive. Additionally, the vertex form or intercepts can be used to verify the correct function. Based on this, the function that matches the graph is likely **C**: \( f(x) = 2x^2 + 4x + 1 \).
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