How do you graph #y= -x^2 + 5#?

Question

Nathan Axel · Accepted Answer

By assigning different values to x, find y and then produce graph

If x is zero, #y=5# If x is 1, #y=4# etc. graph{5 - x^2 [-9.83, 10.17, -4.52, 5.48]}

Eva Adamson · Answer

undefined To graph the quadratic function y = -x^2 + 5, we can start by identifying key features such as the vertex, axis of symmetry, and intercepts.

1. Vertex: The vertex of the parabola represented by the function y = ax^2 + bx + c is given by the coordinates (-b/2a, f(-b/2a)), where f(x) represents the function. In this case, a = -1, b = 0, and c = 5. Therefore, the x-coordinate of the vertex is -0/(2*-1) = 0, and the y-coordinate is f(0) = -0^2 + 5 = 5. So, the vertex is at (0, 5).

2. Axis of Symmetry: The axis of symmetry of a parabola is a vertical line that passes through its vertex. In this case, the axis of symmetry is the vertical line x = 0.

3. Intercepts: To find the y-intercept, set x = 0 in the equation y = -x^2 + 5. So, y = -0^2 + 5 = 5. Therefore, the y-intercept is (0, 5). To find the x-intercepts, set y = 0 and solve for x. 0 = -x^2 + 5 leads to x^2 = 5. Taking the square root of both sides gives x = ±√5. So, the x-intercepts are (√5, 0) and (-√5, 0).

4. Symmetry: The graph of y = -x^2 + 5 is symmetric about its axis of symmetry, which is the y-axis in this case.

Now, with this information, we can sketch the graph of y = -x^2 + 5. The parabola opens downwards since the coefficient of x^2 is negative. The vertex is at (0, 5), and the y-intercept is also at (0, 5). The x-intercepts are at (√5, 0) and (-√5, 0). The axis of symmetry is the y-axis.

Therefore, the graph of y = -x^2 + 5 is a downward-opening parabola with its vertex at (0, 5), intercepting the y-axis at (0, 5), and having x-intercepts at (√5, 0) and (-√5, 0).