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Medieval Archer

Alignments to Content Standards: F-BF.B.3


A computer game uses functions to simulate the paths of an archer’s arrows. The $x$-axis represents the level ground on which the archer stands, and the coordinate pair $(2, 5)$ represents the top of a castle wall over which he is trying to fire an arrow.

In response to user input, the first arrow followed a path defined by the function $f(x) = 6 - x^2$, failing to clear the castle wall.


The next arrow must be launched with the same force and trajectory, so the user must reposition the archer in order for his next arrow to have any chance of clearing the wall.

  1. How much closer to the wall must the archer stand in order for the arrow to clear the wall by the greatest possible distance?
  2. What function must the user enter in order to accomplish this?
  3. If the user can only enter functions of the form $f(x + k)$, what are all the values of $k$ that would result in the arrow clearing the castle wall?

IM Commentary

This task addresses the first part of standard F-BF.3: “Identify the effect on the graph of replacing $f(x)$ by $f(x) + k$, $kf(x)$, $f(kx)$, and $f(x + k)$ for specific values of $k$ (both positive and negative).” Here, students are required to understand the effect of replacing $x$ with $x + k$, but this task can also be modified to test or teach function-building skills involving $f(x) + k$, $kf(x)$, and $f(kx)$ in a similar manner. Solution complexity builds as students progress from part a to part c. While knowing how to solve parts a and b is essential for this standard, understanding the algebraic solution (which ties to A-CED.1) extends beyond common expectations. Instructors therefore may choose to use part c as an enrichment exercise or, alternatively, allow graphical solutions for part c.


  1. The maximum value of $f(x)$ is $6$, which occurs at $x=0$. The maximum value must occur at $x=2$ in order for the arrow to clear the castle wall by the greatest margin. This is accomplished by moving the entire graph two units to the right, essentially moving the archer two units closer to the castle wall.

  2. The user must enter a function whose output is $f(0)$ when the input is $x=2$. A function that subtracts $2$ from every input value before following the procedures of $f$ would accomplish this--namely, $f(x - 2)$.


    Since $f$ is defined as $f(x) = 6 - x^2$, the function $f(x - 2)$ is defined as follows: $$ \begin{align} f(x - 2) &= 6 - (x - 2)^2\\ &= 6 - (x^2 - 4x + 4)\\ &= 6 - x^2 + 4x - 4\\ &= -x^2 + 4x + 2 \end{align} $$
  3. Part b reveals one value ($-2$) of $k$ for which the function $f(x + k)$ would result in the arrow clearing the wall. Setting up and solving an inequality will reveal all such values of $k$.

    First, identify $f(x + k)$ as defined in this particular case:

    $$ \begin{align} f(x + k) &= 6 - (x + k)^2 \\ &=6-x^2 -2kx-k^2 \end{align} $$

    Substituting $2$ for $x$ yields an expression that represents all possible heights of the second arrow when it reaches the castle wall:

    $$ \begin{align} f(2+k) &= 6 - 2^2 -2k(2)-k^2 \\ &= 6 - 4 - 4k - k^2 \\ &= 2 - 4k - k^2 \end{align} $$

    In order for the second arrow to clear the wall, this expression must be greater than $5$. The values of $k$ for which this is true are found by solving the inequality:

    $$ \begin{align} 2 - 4k - k^2 \gt& 5 \\ -3 - 4k - k^2 \gt& 0 \\ k^2 + 4k + 3 \lt& 0 \\ (k + 1)(k + 3) \lt& 0 \end{align} $$

    One of the factors ($k + 1$ or $k + 3$) must be negative, and the other must be positive. The sign of the first factor varies according to three cases: $k + 1$ is negative when $k \lt -1$, zero when $k = -1$, and positive when $k \gt -1$. The sign of the second factor also varies according to three cases: $k + 3$ is negative when $k \lt -3$, zero when $k = -3$, and positive when $k \gt -3$.

    The lists above show that a value for $k$ that is less than $-1$ and greater than $-3$ will make $k + 1$ negative and $k + 3$ positive. Additionally, it shows that a value for $k$ that is greater than $-1$ and less than $-3$ will make $k + 1$ positive and $k + 3$ negative. However, since $k$ cannot simultaneously be greater than $-1$ and less than $-3$, only the former scenario is possible: $k \lt -1$ and $k \gt -3$, or $-3 \lt k \lt -1$.

    Another approach to solving the inequality $(k + 1)(k + 3) \lt 0$ is to recognize that $k + 1$ is necessarily less than $k + 3$, which requires the following to be true if the two factors have opposite signs:

    $$ \begin{align} k+1 \lt &0 \lt k+3 \\ 1 \lt -&k \lt 3 \\ -1 \gt &k \gt -3 \\ -3 \lt &k \lt -1 \end{align} $$

    Therefore the second arrow, when following a path defined by $f(x + k)$, will clear the castle wall if $k$ is greater than $-3$ and less than $-1$.