B. f(x) = { x², x < 1
{ ¹/₃x + ²/₃, x > 1
The function graphed so far has been defined over their domains by a single rule. Some functions, however, are defined by applying different rules at different parts of their domains. These kinds of functions are called piecewise-defined functions.
The Graph A
The graph A is called a parabola with the equation [tex]\boxed{ \ y = a(x - h)^2 + k \ }[/tex] where (h, k) is the vertex (or turning point).
The equation of graph A is [tex]\boxed{ \ y = x^2 \ }[/tex]
The Graph B
The graph B is called a linear function with the equation [tex]\boxed{ \ y = mx + n \ }[/tex].
See the graphs in the attached picture.
In fact, for all x < 0 we have [tex]\boxed{f(x) = x^2}[/tex] and for all x > 0 we have [tex]\boxed{f(x) = \frac{1}{3}x + \frac{2}{3}}[/tex]. That is, the graph of this piecewise-defined function will look like the graph of [tex]\boxed{f(x) = x^2}[/tex] for x < 0 and [tex]\boxed{f(x) = \frac{1}{3}x + \frac{2}{3}}[/tex] for x > 0.
Note that in the above graphs of each rule are not connected to each other. We will name these kinds of functions as discontinuous functions.
[tex]\boxed{ \ f(x) = \left \{ {{x^2, \ \ \ \ \ \ \ \ \ x < 1} \atop {\frac{1}{3}x + \frac{2}{3}, \ \ \ \ x > 1}} \right. \ }[/tex]
Keywords: the function graphed below, consider, which, linear function, gradient, the slope, parabola. vertex, passing through the point, piecewise-defined function, discontinuous, represent
Answer:
Option B.
Step-by-step explanation:
The given represents a piecewise function.
The vertex form of a parabola is
[tex]g(x)=a(x-h)^2+k[/tex]
where, a is constant and (h,k) is vertex.
For x<1, it is a parabola with vertex (0,0). It means the equation of parabola is
[tex]f(x)=ax^2[/tex]
It passes through (1,1), so
[tex]1=a(1)^2[/tex]
[tex]a=1[/tex]
The value of a is 1. So, [tex]f(x)=x^2[/tex] for x<1.
For x>1, the function is a straight line which passes through the point (4,2) and (7,3).
If a line passes through two points [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex], then the equation of line is
[tex]y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)[/tex]
So, the equation of line is
[tex]y-2=\frac{3-2}{7-4}(x-4)[/tex]
[tex]y-2=\frac{1}{3}(x-4)[/tex]
[tex]y=\frac{1}{3}x-\frac{4}{3}+2[/tex]
[tex]y=\frac{1}{3}x+\frac{2}{3}[/tex]
The function is [tex]f(x)=\frac{1}{3}x+\frac{2}{3}[/tex] for x>1.
There is an open circle at x=1. It means function is not defined for x=1.
The required function is
[tex]f(x)=\begin{cases}x^2 & \text{ if } x<1 \\ \frac{1}{3}x+\frac{2}{3} & \text{ if } x>1 \end{cases}[/tex]
Therefore, the correct option is B.
