Graph the quadratic function y = -x2+4x+1. Find the vertex, y-intercept, and x-intercepts.

Step 1: Identify the type of graph and its general shape. The given equation y = -x^2+4x+1 is a quadratic function of the form y = ax^2+bx+c. Since the coefficient of x^2 is a=-1 (which is negative), the graph is a parabola that opens downwards. Step 2: Find the vertex of the parabola. The x-coordinate of the vertex is given by the formula x = -(b)/(2a). For y = -x^2+4x+1, we have a=-1 and b=4. x = -(4)/(2(-1)) = -(4)/(-2) = 2 Substitute x=2 into the equation to find the y-coordinate of the vertex: y = -(2)^2 + 4(2) + 1 = -4 + 8 + 1 = 5 The vertex is at (2, 5). This is the maximum point of the parabola. Step 3: Find the y-intercept. The y-intercept occurs when x=0. y = -(0)^2 + 4(0) + 1 = 1 The y-intercept is at (0, 1). Step 4: Find the x-intercepts (roots). The x-intercepts occur when y=0. -x^2+4x+1 = 0 Multiply by -1 to make the leading coefficient positive: x^2-4x-1 = 0 Use the quadratic formula x = -b ± sqrt(b^2-4ac)2a: x = -(-4) ± sqrt((-4)^2 - 4(1)(-1))2(1) x = 4 ± sqrt(16 + 4)2 x = 4 ± sqrt(20)2 x = 4 ± 2sqrt(5)2 x = 2 ± sqrt(5) The x-intercepts are approximately x_1 = 2 - sqrt(5) ≈ 2 - 2.236 = -0.236 and x_2 = 2 + sqrt(5) ≈ 2 + 2.236 = 4.236. So, the x-intercepts are approximately (-0.24, 0) and (4.24, 0). Step 5: Sketch the graph. To sketch the graph, plot the following points: • Vertex: (2, 5) • Y-intercept: (0, 1) • X-intercepts: (2-sqrt(5), 0) and (2+sqrt(5), 0) The parabola opens downwards, is symmetric about the vertical line x=2 (the axis of symmetry), and passes through these points. Last free one today — make it count tomorrow, or type /upgrade for unlimited.