Answer:
Ix + 2I - 5 = -(x - 1)*(x - 3)
Step-by-step explanation:
In the graph, we can see two functions a red one, f(x), and a blue one, g(x)
And the equation would be f(x) = g(x)
The red one is an absolute value function, we can see that the vertex is at x = -2 and at y = -5, and it has a slope equal to 1.
Remember that for an absolute value equation with a vertex (a, b) is:
f(x) = Ix - aI + b
Then in this case, the absolute value equation is:
f(x) = Ix - (-2)I - 5
f(x) = Ix + 2I - 5
The blue equation is a quadratic equation, with roots 1 and 3, and we can see that it opens down, so the leading coefficient is negative.
Remember that for a quadratic equation with roots a and b, with leading coefficient A is:
g(x) = A*(x - a)*(x - b)
Then in this case the quadratic equation is something like:
g(x) = -N*(x - 1)*(x - 3)
Then the equation will be something like:
f(x) = Ix + 2I - 5 = g(x) = -N*(x - 1)*(x - 3)
Ix + 2I - 5 = -N*(x - 1)*(x - 3)
The only option that has this shape is the first option (the top one) such that N = 1, then:
Ix + 2I - 5 = -(x - 1)*(x - 3)
is the correct option.
