Starting with the given quadratic equation, y=x(x-4), we first need to identify the form of this equation which is y=ax^2+bx+c. Comparing this to our initial equation, we can identify a=1, b=-4, and c=0.
To identify the vertex of the equation, we have to first find the value of h which is -b/2a. Substituting the values of a and b in this equation, we find h= 4/2 = 2.
Next, we will calculate the value of y for the identified h using the function f(h) = 2(2-4) = -4. Hence, the vertex of the equation is at the coordinates (2,-4).
We also can find that the axis of symmetry is the line x=h which equals x=2.
The value of a, which is equal to 1, is positive. Therefore, our parabola will open upwards indicating that it has a minimum value. The minimum value in this case is the y-coordinate of our vertex, which is -4.
To find where the graph crosses the y-axis, we have to find the y-intercept. This happens when x=0. So, we can plug in x=0 into our equation to get y = 0*(0-4) = 0. Consequently, the y-intercept is the point where the graph crosses the y-axis at (0,0).
Finally, we need to find the x-intercepts, which are the points where the graph crosses the x-axis. These points can be found when y=0. Solving the equation x(x-4) = 0, we get values for x as 0 or 4. So, the x-intercepts are the points (0,0) and (4,0).
To summarize we have:
Vertex: (2, -4)
Axis of symmetry: x=2
Minimum value: -4
Y-intercept: (0, 0)
X-intercepts: (0, 0) and (4, 0)