Piecewise Intersection

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equations

Below are two piecewise defined functions and their graphs: $V(h)$ and $K(m)$ .

$V(h) = \begin {cases} -2h-3 & \text { if } [-6, -2] \\ -(h+3)(h-3) & \text { if } (-2, 4] \\ \frac {7h}{4} - 8 & \text { if } (4,8) \end {cases}$

$K(m) = \begin {cases} m+4 & \text { if } (-7, 0] \\ m^2-7m+9 & \text { if } (1,6] \end {cases}$

Find the points of intersection between the graphs of $V$ and $K$ .

From the graphs there appear to be three points of intersection.

First Point

We’ll set the function formulas equal to each other and solve the equation formed.

$-2x-3 = \answer {x+4}$

$\answer {-3x} = 7$

$x = \frac {7}{-3} = -\frac {7}{3}$

Check....

$V(-\tfrac {7}{3}) = -2 \cdot -\tfrac {7}{3} - 3 = \tfrac {14}{3} - \tfrac {9}{3} = \tfrac {5}{3}$
$K(-\tfrac {7}{3}) = -\tfrac {7}{3} + 4 = -\tfrac {7}{3} + \frac {12}{3} = \tfrac {5}{3}$

The intersection point is $\left ( -\frac {7}{3}, \frac {5}{3} \right )$ .

Second Point

We’ll set the function formulas equal to each other and solve the equation formed.

$\begin{align*} \answer{-(t+3)(t-3)} &= t^2 - 7 t + 9 \\ \answer{-t^2 + 9} &= t^2 - 7 t + 9 \\ 0 &= 2 t^2 - 7t \\ 0 &= t(2t - 7) \end{align*}$

This gives two solutions, $0$ and $\frac {7}{2}$ .

We can see from the graph that the full graphs of $-(t+3)(t-3)$ and $t^2 - 7 t + 9$ would intersect at $t = 0$ as well. However, our piecewise defined functions do not share these formulas at $0$ . The definitions of the piecewise functions do not allow $0$ as a common solution.

After considering the domains of $V$ and $H$ , the only solution here is $\frac {7}{2}$ .

Check....

$V(\tfrac {7}{2}) = -\left (\tfrac {7}{2} + 3\right ) \left (\tfrac {7}{2} - 3\right ) = -\tfrac {13}{2} \cdot \tfrac {1}{2} = -\frac {13}{4}$
$K(\tfrac {7}{2}) = \left (\tfrac {7}{2}\right )^2 - 7\left (\tfrac {7}{2}\right ) + 9 = -\frac {13}{4}$

The intersection point is $\left (\tfrac {7}{2}, -\frac {13}{4}\right )$ .

Third Point

We’ll set the function formulas equal to each other and solve the equation formed.

$\begin{align*} \answer{\frac{7w}{4} - 8} &= w^2 - 7 w + 9 \\ 0 &= w^2 - \answer{\frac{35w}{4}} + 17 \\ 0 &= \frac{-(- \tfrac{35}{4}) \pm \sqrt{(- \tfrac{35}{4})^2 - 4 \cdot 1 \cdot 17}}{2 \cdot 1} t \\ 0 &= \frac{\tfrac{35}{4} \pm \sqrt{\tfrac{137}{16}}}{2} \\ 0 &= \frac{35}{8} \pm \frac{\sqrt{137}}{8} \\ 0 &= \frac{35 \pm \sqrt{137}}{8} \end{align*}$

The graph shows us that we only have one of these solutions because the line segment doesn’t extend further to the left to intersect the parabola twice.

The only soution here is $w = \frac {35 + \sqrt {137}}{8}$ .

Check....

$V\left (\frac {35 + \sqrt {137}}{8}\right ) = \frac {1}{32}(7 \sqrt {137} - 11)$
$K\left (\frac {35 + \sqrt {137}}{8}\right ) = \frac {1}{32}(7 \sqrt {137} - 11)$

The intersection point is $\left (\frac {35 + \sqrt {137}}{8}, \answer {\frac {1}{32}(7 \sqrt {137} - 11)} \right )$ .

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more examples can be found by following this link
More Examples of Piecewise-Defined Functions