Consider the function graphed below Which function does this graph represent?A. f(x) = { x^2,x<1 3x - 2,x>1B. f(x) = { x^2,x<1 1/3x + 2/3,x>1C. f(x) = { x^2,x<1 3x, x>1D. f(x) = { x^2,x<1 1/3x - 2,x>1
Accepted Solution
A:
B. f(x) = { x², x < 1 { ¹/₃x + ²/₃, x > 1Further explanationThe 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 AThe 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).[tex](h. k) \rightarrow y = a(x - 0)^2 + 0 \rightarrow \boxed{ \ y = ax^2 \ }[/tex]Passing through the point (1, 1)[tex](1, 1) \rightarrow y = ax^2 \rightarrow 1 = a(1)^2 \rightarrow \boxed{ \ a = 1 \ }[/tex]The equation of graph A is [tex]\boxed{ \ y = x^2 \ }[/tex]The Graph BThe graph B is called a linear function with the equation [tex]\boxed{ \ y = mx + n \ }[/tex].Passing through (1, 1) and (4, 2).The slope or gradient [tex]\boxed{ \ m = \frac{y_2 - y_1}{x_2 - x_1} \ } \rightarrow \boxed{ \ m = \frac{2 - 1}{4 - 1} = \frac{1}{3} \ }[/tex]Thus becoming [tex]\boxed{ \ y = \frac{1}{3}x + n\ }[/tex][tex]\boxed{ \ (1, 1) \ \rightarrow y = \frac{1}{3}x + n} \rightarrow 1 = \frac{1}{3}(1) + n \rightarrow \boxed{ \ n = \frac{2}{3} \ }[/tex]The equation of graph B is [tex]\boxed{ \ y = \frac{1}{3}x + \frac{2}{3} \ }[/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]Learn moreDetermine the midpoint of two endpoints line equation as a slope-intercept form line that is not parallel to either the x-axis or the y-axis : the function graphed below, consider, which, linear function, gradient, the slope, parabola. vertex, passing through the point, piecewise-defined function, discontinuous, represent