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
Real Tutor Solution
Answer
Solution
Reviewed and approved by the UpStudy tutoring team
Extra Insights
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.