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Given \( f(x)=-3 x^{2}+3 \), sketch the graph of \( f \), clearly showing ALL intercepts with the axes and turning points. 1. Use your sketch to answer the following. INSTRUCTION: Give your answers as coordinates in the form \( (\mathbf{x} ; \mathbf{y}) \). Answer: 1. The \( x \)-intercepts of \( f \) are and 2. The \( y \)-intercept of \( f \) is 3. The graph of \( f \) is Select

Ask by Hampton Kelley. in South Africa
Jan 24,2025

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The \( x \)-intercepts are \( (1, 0) \) and \( (-1, 0) \), the \( y \)-intercept is \( (0, 3) \), and the graph is a downward-opening parabola with its vertex at \( (0, 3) \).

Solution

To analyze the function \( f(x) = -3x^2 + 3 \), we will follow these steps: 1. **Find the intercepts with the axes.** 2. **Determine the turning points.** 3. **Sketch the graph based on the findings.** ### Step 1: Find the intercepts **1.1. Find the \( x \)-intercepts:** The \( x \)-intercepts occur when \( f(x) = 0 \). \[ -3x^2 + 3 = 0 \] Rearranging gives: \[ 3x^2 = 3 \] Dividing both sides by 3: \[ x^2 = 1 \] Taking the square root of both sides: \[ x = \pm 1 \] Thus, the \( x \)-intercepts are \( (1; 0) \) and \( (-1; 0) \). **1.2. Find the \( y \)-intercept:** The \( y \)-intercept occurs when \( x = 0 \). \[ f(0) = -3(0)^2 + 3 = 3 \] Thus, the \( y \)-intercept is \( (0; 3) \). ### Step 2: Determine the turning points The function \( f(x) = -3x^2 + 3 \) is a quadratic function that opens downwards (since the coefficient of \( x^2 \) is negative). The vertex of the parabola, which is the turning point, can be found using the formula: \[ x = -\frac{b}{2a} \] For our function, \( a = -3 \) and \( b = 0 \): \[ x = -\frac{0}{2 \cdot -3} = 0 \] Now, substituting \( x = 0 \) back into the function to find the \( y \)-coordinate of the vertex: \[ f(0) = 3 \] Thus, the turning point (vertex) is \( (0; 3) \). ### Step 3: Sketch the graph Now we can summarize the findings: - \( x \)-intercepts: \( (1; 0) \) and \( (-1; 0) \) - \( y \)-intercept: \( (0; 3) \) - Turning point: \( (0; 3) \) ### Final Answers 1. The \( x \)-intercepts of \( f \) are \( (1; 0) \) and \( (-1; 0) \). 2. The \( y \)-intercept of \( f \) is \( (0; 3) \). 3. The graph of \( f \) is a downward-opening parabola with its vertex at the \( y \)-intercept. The graph will look like a parabola opening downwards, intersecting the x-axis at \( (1; 0) \) and \( (-1; 0) \), and the y-axis at \( (0; 3) \).

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To sketch the graph of the function \( f(x) = -3x^2 + 3 \), let's first identify its key features. 1. The \( x \)-intercepts occur when \( f(x) = 0 \): \[ -3x^2 + 3 = 0 \implies -3x^2 = -3 \implies x^2 = 1 \implies x = \pm 1 \] Therefore, the \( x \)-intercepts are \( (1; 0) \) and \( (-1; 0) \). 2. The \( y \)-intercept occurs when \( x = 0 \): \[ f(0) = -3(0)^2 + 3 = 3 \] Therefore, the \( y \)-intercept is \( (0; 3) \). 3. The graph of \( f \) is a downward-opening parabola since the coefficient of \( x^2 \) is negative. The turning point, or vertex, can be found using the formula \( x = -\frac{b}{2a} \), where \( a = -3 \) and \( b = 0 \). This gives the vertex at \( (0; 3) \), making it the highest point on the graph. Now, let's fill in the answers: 1. The \( x \)-intercepts of \( f \) are \( (1; 0) \) and \( (-1; 0) \). 2. The \( y \)-intercept of \( f \) is \( (0; 3) \). 3. The graph of \( f \) is a downward-opening parabola.

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