The companion set to Workshop 11. Twelve SAT-style practice questions for solving and checking with the built-in Desmos calculator: solve by seeing, slide a constant, match the curve, and read a graph in context. Translate, type, read.
Every question here is one the built-in Desmos calculator can show you.
Translate: for an equation, graph both sides and look for where they meet (or just touch). For two linear equations, no solution is two parallel lines and infinitely many is one line on top of another. Equivalent means the same curve; an inequality is a shaded region.
Type: enter what you translated. Add a slider when a constant is unknown; tap its value to type a fraction or a number past 10. Compare answer choices one at a time (tap a row's colored icon to hide it).
Read: tap crossings and vertices, and read circle edges from the grid. Make sure you report the number the question asks for.
Check: confirm with a quick bit of algebra.
Here is how the wrong answers try to trip you:
Extraneous Root. Keeps an algebra answer that makes a denominator zero. Graph both sides and count the crossings.
Other Crossing. Reports the wrong crossing when the question asks for the greater (or lesser) solution.
Time, Not Height. Reports the x-coordinate of a point when the question asks for its y-coordinate, or the reverse.
Inside, Not Outside. Shifts sideways when the question says up or down. Outside the function moves it vertically.
r Squared, Not r. Reports \(r^2\) instead of \(r\). Read the circle's width off the graph and halve it.
Sign Slip. Flips a sign while moving terms by hand. Typing the equation exactly as given avoids it.
Question 1 of 12
Solve by SeeingMedium
$(x+1)(x-9) = -21$
What is the greater solution to the given equation?
The takeaway:
Translate: a solution is an $x$ where the two sides are equal, so graph each side.
Type: $y = (x+1)(x-9)$ and $y = -21$, then zoom out until the line meets the parabola.
Read: tap the crossings, $(2, -21)$ and $(6, -21)$. The greater solution is $x = 6$.
Check: $x^2 - 8x + 12 = 0$ factors as $(x - 6)(x - 2) = 0$.
Match the CurveMedium
$P(x) = 8x^4 - 12x^3$
Which of the following is equivalent to $P(x)$?
The takeaway:
Translate: equivalent means the same graph for every $x$.
Type: $y = 8x^4 - 12x^3$, then each choice as its own $y = \dots$ row. Tap a row's colored icon to hide it, so only the original and one choice show at a time.
Read: only choice C lands exactly on top of the original and covers it. Each of the others peels away somewhere. An overlap suggests a match; the check below confirms it for every $x$.
Check: the greatest common factor is $4x^3$, and $8x^4 - 12x^3 = 4x^3(2x - 3)$.
TransformationsMedium
The function $f$ is defined by $f(x) = 6x - 5$.
The function $g$ is defined so that $y = g(x)$ is the result of shifting $y = f(x)$ down 8 units.
Which equation defines $g$?
The takeaway:
Translate: shifting down 8 means $g(x) = f(x) - 8$. Outside the function moves it up or down; inside moves it sideways.
Type: $f(x) = 6x - 5$, then $y = f(x) - 8$, then the right side of each choice as its own row, like $y = 6x + 3$.
Read: choice D, $y = 6x - 13$, lands exactly on $y = f(x) - 8$.
Check: $6x - 5 - 8 = 6x - 13$.
Absolute ValueMedium
$|4x + 8| \ge 20$
Which of the following is not a solution to the given inequality?
The takeaway:
Translate: nothing to change; Desmos reads absolute value directly.
Type: $|4x + 8| \ge 20$ exactly as written.
Read: Desmos shades two regions, $x \le -7$ and $x \ge 3$. Their edges are solid, so $-7$ and $3$ are solutions. The unshaded gap holds the values that are not solutions, and $1$ is in the gap.
Check: $|4(1) + 8| = 12$, and $12 \ge 20$ is false.
Slide a ConstantChallenge
$4(x - 2y) = 3x + 12$
$5x + ky = 2(3x - 6) + y$
In the given system of equations, $k$ is a constant. If the system has infinitely many solutions, what is the value of $k$?
The takeaway:
Translate: infinitely many solutions means the two equations draw the same line.
Type: both equations exactly as given. Desmos offers a slider for $k$; add it.
Read: drag $k$ until the second line lands on the first (near $k = 9$), then type $k = 9$ exactly: the lines match.
Check: the first equation is $x - 8y = 12$, and the second becomes $x - (k - 1)y = 12$, so $k - 1 = 8$.
Solve by SeeingChallenge
$x^2 + 4x + y^2 - 20y = 40$
The given equation defines a circle in the $xy$-plane. What is the length of the circle's radius?
The takeaway:
Translate: nothing to change; Desmos graphs the circle as written.
Type: $x^2 + 4x + y^2 - 20y = 40$, then zoom out until the whole circle shows.
Read: on the grid, the circle runs from $y = -2$ at the bottom to $y = 22$ at the top (add the lines $y = -2$ and $y = 22$ to confirm each just touches it). That height is a diameter, $22 - (-2) = 24$, so the radius is $12$.
Check: completing the square gives $(x + 2)^2 + (y - 10)^2 = 144$, and $\sqrt{144} = 12$.
Slide a ConstantChallenge
$2(4x + y) = 3x - y + 9$
$kx + 6y = 4$
In the given system of equations, $k$ is a constant. If the system has no solution, what is the value of $k$?
The takeaway:
Translate: no solution means two parallel lines that never meet.
Type: both equations exactly as given, and add the slider for $k$.
Read: the slider starts at $-10$ to $10$; drag $k$ to the right end, $k = 10$ (or type $10$). The second line runs parallel to the first, and the check below confirms the lines never meet. (Or tap the $k$ value and type each choice: $-10$, $5$, and $24$ all leave a crossing.)
Check: the first equation simplifies to $5x + 3y = 9$. Doubling gives $10x + 6y = 18$, and $4 \ne 18$, so the lines are distinct.
The takeaway:
Translate: graph each side as its own $y = \dots$ and look for crossings.
Type: $y = \frac{x}{x-3} + \frac{2}{x+3}$ and $y = \frac{18}{x^2 - 9}$.
Read: the curves cross once, just above the x-axis; tap the crossing (zoom in if needed) to see $(-8, 0.327)$, so $x = -8$. Both sides are undefined at $x = 3$ (a vertical asymptote), so there is no crossing there.
Check: the equation requires $x \ne 3$ and $x \ne -3$. Clearing denominators gives $(x + 8)(x - 3) = 0$, and $x = 3$ is not allowed, so it is extraneous.
Match the CurveChallenge
$(x^2 + kx - 2)(3x - 4) = 3x^3 + 11x^2 - 26x + 8$
In the given equation, $k$ is a constant. If the equation is true for all values of $x$, what is the value of $k$?
The takeaway:
Translate: true for all values of $x$ means the two sides draw the same curve.
Type: $y = 3x^3 + 11x^2 - 26x + 8$, then $y = (x^2 + kx - 2)(3x - 4)$, and add the slider for $k$.
Read: drag $k$ until the curves match (near $k = 5$), then type $k = 5$ exactly: the second curve lands on the first. (Both curves pass through $(0, 8)$ and $(\frac{4}{3}, 0)$ for every $k$, so look for the whole curves to overlap, not just those two points.)
Check: the $x^2$ terms give $3k - 4 = 11$, so $k = 5$, and the $x$ terms agree: $-4k - 6 = -26$.
Slide a ConstantChallenge
$px^2 + 20x + 8 = 0$
In the given equation, $p$ is a positive constant. The equation has exactly one distinct real solution. What is the value of $p$?
The takeaway:
Translate: exactly one solution means the parabola $y = px^2 + 20x + 8$ touches the x-axis exactly once.
Type: $y = px^2 + 20x + 8$ and add the slider for $p$. The slider runs from $-10$ to $10$, but you can tap its value and type each choice directly: $10$, $50$, $25$, $\frac{25}{2}$.
Read: $p = 10$ crosses the x-axis twice, $p = 25$ and $p = 50$ miss it, and $p = \frac{25}{2}$ just touches it, at $(-0.8, 0)$.
Check: the discriminant $20^2 - 4(p)(8) = 400 - 32p$ is $0$ when $p = \frac{25}{2}$.
Solve by SeeingChallenge
$y \le 5x + 3$
$y \ge -x + 4$
Which point $(x, y)$ is a solution to the given system of inequalities in the xy-plane?
The takeaway:
Translate: a solution to the system is a point where both shaded regions overlap.
Type: $y \le 5x + 3$ and $y \ge -x + 4$, then the four points on one line: $(1, 1), (1, 5), (0, 10), (2, -1)$.
Read: only $(1, 5)$ sits in the overlap. Each of the other points is inside just one of the two shaded regions.
Check: $5 \le 8$ and $5 \ge 3$ are both true.
In ContextChallenge
An object is launched upward from a platform. Its height, in feet, $t$ seconds after launch is modeled by the function below, where $c$ is a constant.
$h(t) = -16t^2 + 64t + c$
The object is $100$ feet above the ground $1$ second after launch. According to the model, what is the object's maximum height, in feet?
The takeaway:
Translate: time is $x$ and height is $y$. The object is at $(1, 100)$, so pin $c$ first: $-16 + 64 + c = 100$ gives $c = 52$.
Type: $y = -16x^2 + 64x + 52$, then set the window (wrench) to about $-1 \le x \le 6$ and $-10 \le y \le 130$ so the whole flight fits.
Read: tap the top of the curve. The vertex is $(2, 116)$, so the maximum height is the y-coordinate, $116$ feet, reached at $t = 2$ seconds.
Check: $h(2) = -64 + 128 + 52 = 116$.
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