Question

The graph represents the piecewise function

Answer 1

Piecewise function can be explained as a set of subfunctions (each subfunction applied to an interval). Here we can observe a piecewise function that is composed of three sub functions.

-The first subfunction: between 0 and 3 (open interval in 3) there is a quadratic function, we can obtain it from the general form of the quadratic function when it has a vertex on (0,0)  which is:

$f(x)=ax^{2}$

$a$ will be a positive value if the parabola is opening upward and negative if it is opneing downward. This one opens upward so it is positive and it will be =1 because when we make $x=1$  on the graph corresponds to $y=1$ so:

$f(x)=ax^{2}\\1=a(1^{2})\\a=1$

Our subfunction if $0\leq x$ is $f(x)=x^{2}$

-the second and third subfunctions: both are straight lines defined each in an interval. The first between 3 and 6 (open interval in 3) and the second between 6 and 10. To build this two functions we need to understand the general form of the line function:

$f(x)=mx+b$

where m is the slope of the line and b is the y intercept.

we calculate first m and b for both lines with the m and b equation (we must take two coordintates of each line ($x_{1},y_{1},x_{2},y_{2}$):

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$b=y-mx$

First line (4,6),(5,5)

$m=\frac{5-6}{5-4}}=-1$

$b=(5)-(-1)(5)=10$

Second line (8,7),(10,10)

$m=\frac{10-7}{10-8}}=\frac{3}{2}}=1.5$

$b=(10)-(\frac{3}{2}})(10)=-5$

Now we can build both the equations:

First Lline ($3$)

$f(x)=mx+b\\f(x)=(-1)x+10\\f(x)=10-x$

Second Line($6\leq x\leq 10$):

$f(x)=mx+b\\f(x)=(\frac{3}{2}})x-5$

Answer 2

Drop-down menu selections are not provided.

f(x) = {x^2, 0 ≤ x < 3
.. .. .. {10 -x, 3 < x ≤ 6
.. .. .. {(3/2)x -5, 6 ≤ x ≤ 10

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Note that the last two function segments both give (6, 4), but it is bad form to define a piecewise function such that the pieces overlap. That is, f(6) has two definitions.