# How do you find its vertex, axis of symmetry, y-intercept and x-intercept for #f(x) = x^2 - 4x#?

Vertex is

graph{(x^2-4x-y)(x-2)((x-2)^2+(y+4)^2-0.03)=0 [-9.92, 10.08, -5.12, 4.88]}

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To find the vertex of a quadratic function in the form f(x) = ax^2 + bx + c, you use the formula: Vertex = (-b/2a, f(-b/2a)). To find the axis of symmetry, use the formula: Axis of Symmetry = -b/2a. The y-intercept is found by setting x = 0 and solving for y: f(0) = c. To find the x-intercepts (or roots), set f(x) = 0 and solve for x using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a).

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