Question

graph the equation $y = -x^2 + 2x + 3$ on the accompanying set of axes. you must plot 5 points including the roots and the vertex. using the graph, determine the equation of the axis of symmetry. click to plot points. click points to delete them.

Explanation:

Step1: Find the roots (x-intercepts)

Set \( y = 0 \), so \( -x^2 + 2x + 3 = 0 \). Multiply both sides by -1: \( x^2 - 2x - 3 = 0 \). Factor: \( (x - 3)(x + 1) = 0 \). Thus, roots are \( x = 3 \) (when \( x - 3 = 0 \)) and \( x = -1 \) (when \( x + 1 = 0 \)). So points are \( (-1, 0) \) and \( (3, 0) \).

Step2: Find the vertex

For a quadratic \( y = ax^2 + bx + c \), the x-coordinate of the vertex is \( x = -\frac{b}{2a} \). Here, \( a = -1 \), \( b = 2 \), so \( x = -\frac{2}{2(-1)} = 1 \). Substitute \( x = 1 \) into the equation: \( y = -(1)^2 + 2(1) + 3 = -1 + 2 + 3 = 4 \). So vertex is \( (1, 4) \).

Step3: Find two more points

Choose \( x = 0 \): \( y = -0 + 0 + 3 = 3 \), so point \( (0, 3) \). Choose \( x = 2 \): \( y = -(4) + 4 + 3 = 3 \), so point \( (2, 3) \).

Step4: Determine the axis of symmetry

The axis of symmetry of a parabola is the vertical line through the vertex. Since the vertex has x-coordinate 1, the equation is \( x = 1 \).

Answer:

The equation of the axis of symmetry is \( x = 1 \)