Mathematics, 19.09.2019 18:50 Webber07

Consider the function graphed below
which function does this graph represent?
a. f(x) = { x^2,x< 1
3x - 2,x> 1
b. f(x) = { x^2,x< 1
1/3x + 2/3,x> 1
c. f(x) = { x^2,x< 1
3x, x> 1
d. f(x) = { x^2,x< 1
1/3x - 2,x> 1

Consider the function graphed below which function does this graph represent?

Answers: 2

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Answers

Answer from: granniejo

B. f(x) = { x², x < 1

{ ¹/₃x + ²/₃, x > 1

Further explanation

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 $\boxed{ \ y = a(x - h)^2 + k \ }$ where (h, k) is the vertex (or turning point).

$(h. k) \rightarrow y = a(x - 0)^2 + 0 \rightarrow \boxed{ \ y = ax^2 \ }$ Passing through the point (1, 1) $(1, 1) \rightarrow y = ax^2 \rightarrow 1 = a(1)^2 \rightarrow \boxed{ \ a = 1 \ }$

The equation of graph A is $\boxed{ \ y = x^2 \ }$

The Graph B

The graph B is called a linear function with the equation $\boxed{ \ y = mx + n \ }$ .

Passing through (1, 1) and (4, 2).The slope or gradient $\boxed{ \ m = \frac{y_2 - y_1}{x_2 - x_1} \ } \rightarrow \boxed{ \ m = \frac{2 - 1}{4 - 1} = \frac{1}{3} \ }$ Thus becoming $\boxed{ \ y = \frac{1}{3}x + n\ }$ $\boxed{ \ (1, 1) \ \rightarrow y = \frac{1}{3}x + n} \rightarrow 1 = \frac{1}{3}(1) + n \rightarrow \boxed{ \ n = \frac{2}{3} \ }$ The equation of graph B is $\boxed{ \ y = \frac{1}{3}x + \frac{2}{3} \ }$

See the graphs in the attached picture.

In fact, for all x < 0 we have $\boxed{f(x) = x^2}$ and for all x > 0 we have $\boxed{f(x) = \frac{1}{3}x + \frac{2}{3}}$ . That is, the graph of this piecewise-defined function will look like the graph of $\boxed{f(x) = x^2}$ for x < 0 and $\boxed{f(x) = \frac{1}{3}x + \frac{2}{3}}$ 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.

$\boxed{ \ f(x) = \left \{ {{x^2, \ \ \ \ \ \ \ \ \ x < 1} \atop {\frac{1}{3}x + \frac{2}{3}, \ \ \ \ x 1}} \right. \ }$

Learn moreDetermine the midpoint of two endpoints linkThe line equation as a slope-intercept form linkA line that is not parallel to either the x-axis or the y-axis link

Keywords: the function graphed below, consider, which, linear function, gradient, the slope, parabola. vertex, passing through the point, piecewise-defined function, discontinuous, represent

$Consider the function graphed below which function does this graph represent? a. f(x) = { x^2,x<$

Answer from: eve2055

Option B.

Step-by-step explanation:

The given represents a piecewise function.

The vertex form of a parabola is

g(x)=a(x-h)^2+k

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

f(x)=ax^2

It passes through (1,1), so

1=a(1)^2

a=1

The value of a is 1. So, f(x)=x^2 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 (x_1,y_1) and (x_2,y_2) , then the equation of line is

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

So, the equation of line is

$y-2=\frac{3-2}{7-4}(x-4)$

$y-2=\frac{1}{3}(x-4)$

$y=\frac{1}{3}x-\frac{4}{3}+2$

$y=\frac{1}{3}x+\frac{2}{3}$

The function is $f(x)=\frac{1}{3}x+\frac{2}{3}$ for x>1.

There is an open circle at x=1. It means function is not defined for x=1.

The required function is

$f(x)=\begin{cases}x^2 & \text{ if } x1 \end{cases}$

Therefore, the correct option is B.

Answer from: Quest

step-by-step explanation:

Answer from: Quest

i believe the correct answer is b.

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