The equation becomes:
f(x) = -x(x - 2) = -x^2 + 2x
If the tangents to graph are positive for x < 1, then that just means the parabola is increasing from (-infinity, 1). For it to be increasing over that interval, it needs to be upside down, and so I placed a negative sign in front of the leading coefficient, which reflects it in the x-axis. Secondly, if the graph has a vertex at x = 1, then that implies this:
f'(1) = 0 --> Mean the value of the first derivative at x = 1 is 0. Let's see if that's the case.
f'(x) = -2x + 2 --> f'(1) = -2(1) + 2 = 0 --> It works! But we need to know what the y-value of the vertex is at x = 1. The derivative tells us that there definitely is a vertex at x = 1, now to find the y-value of this vertex, we just plug in x = 1 in to our function and get the value:
f(1) = -(1)(1 - 2) = -(1)(-1) = 1 --> So the vertex is at (1, 1)
So this equations gives us that there is definitely a vertex at x = 1, moreover, at (1, 1), it's increasing for x < 1, and it's zeroes are at x = 0 and x = 2. But because we need to graph it, why not also find the y-intercept so it's easier? And we know that the y-intercept occurs when x = 0. So we plug in 0 for x in the function and get y = 0. So the y-intercept occurs at y = 0. This means that both our x and y intercepts occur at (0, 0). So that's everything. Now we just graph with the information we have.
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@ephilo