Linear and Non-Linear Relations: Test Review

Key facts

Where marks get lost

• ×Using first differences when the $x$-values do not go up by equal steps, and then calling a linear relation non-linear (as with $x$: $0,1,3,4$ and $y=2x+1$). (Linear and Non-Linear Relations)
• ×Subtracting in the wrong order (current minus next). For $y$: $20,17,14$ that gives $+3$ instead of $-3$ and makes a shrinking pattern look like a growing one. (Linear and Non-Linear Relations)
• ×Thinking a decreasing relation cannot be linear. A constant negative first difference, like −3, −3, −3, is still linear: the line falls instead of rising. (Linear and Non-Linear Relations)
• ×Deciding from only two points. Any two points can be joined by a straight line, so you need at least three points (and every difference in the table) to judge. (Linear and Non-Linear Relations)
• ×Adding a first difference to predict far ahead in a non-linear pattern. The doubling pattern 100, 200, 400 is not 100 more each step, so adding 100 per step gives badly wrong predictions. (Linear and Non-Linear Relations)
• ×Swapping the two numbers: writing $C=4d+2$ for "4 dollars to start plus 2 dollars per km". The number that multiplies the variable is the rate (2), not the starting fee. (Rate of Change and Initial Value)
• ×Using the change in $y$ as the rate when $x$ goes up by more than 1. In a table with $t=0,5,10$ and $V$ rising by 125 each row, the rate is $125÷5=25$ L per minute, not 125. (Rate of Change and Initial Value)
• ×Taking the first $y$-value in a table as the initial value when the first row is not $x=0$. Work backward to $x=0$ using the rate. (Rate of Change and Initial Value)
• ×Assuming that doubling the independent variable doubles the dependent variable in a partial variation. That only works when the initial value is 0. (Rate of Change and Initial Value)
• ×Dropping the negative sign for a decreasing quantity, writing $V=40t+1200$ for a tub that is draining. (Rate of Change and Initial Value)
• ×Dividing run by rise. For $(1,4)$ and $(5,6)$ the slope is $\frac{6-4}{5-1}=\frac{2}{4}=\frac{1}{2}$, not $\frac{4}{2}=2$. (Slope, y-Intercept and y = mx + b)
• ×Subtracting in different orders on top and bottom, such as $\frac{y_2-y_1}{x_1-x_2}$, which gives the right size but the wrong sign. Watch negatives too: $3-(-1)$ is $4$, not $2$. (Slope, y-Intercept and y = mx + b)
• ×Reading $b$ from where the line crosses the $x$-axis instead of the $y$-axis. (Slope, y-Intercept and y = mx + b)
• ×Saying a vertical line has slope 0. A horizontal line has slope 0; a vertical line's slope is undefined because its run is 0. (Slope, y-Intercept and y = mx + b)
• ×Counting grid squares without checking the scale. If each square on the y-axis is 5 units, a rise of 2 squares is 10, not 2. (Slope, y-Intercept and y = mx + b)
• ×Stopping at the intersection. 'They are equal at 5 GB' does not answer 'which plan should I choose?' Say which plan is cheaper below and above that value. (Comparing Two Linear Relations)
• ×Setting only the rates or only the starting amounts equal (like $5=2$ or $30=45$) instead of setting the whole expressions equal, $5g+30=2g+45$. (Comparing Two Linear Relations)
• ×Solving for $g$ and forgetting to substitute back to find the shared cost, the second coordinate of the intersection. (Comparing Two Linear Relations)
• ×Assuming the option with the smaller starting amount is always cheaper. Plan A starts lower but loses after 5 GB because its rate is higher. (Comparing Two Linear Relations)
• ×Thinking that no solution means you made an error. Parallel lines (same rate, different starting amounts) never meet, so the equation has no solution. (Comparing Two Linear Relations)
• ×Mixing up $x=k$ and $y=k$. Remember that $x=3$ is the set of points whose $x$-coordinate is 3, which form a vertical line. (Special Lines, xy = k and Inequalities)
• ×Giving the $y$-intercept of $x-y=3$ as $3$. Rearranged, it is $y=x-3$, so the $y$-intercept is $-3$. (Special Lines, xy = k and Inequalities)
• ×Drawing a solid boundary for $<$ or $>$. Points on the line do not satisfy a strict inequality, so the line must be dashed. (Special Lines, xy = k and Inequalities)
• ×Using $(0,0)$ as the test point when the boundary passes through the origin, as $y=2x$ does. Pick a point clearly off the line. (Special Lines, xy = k and Inequalities)
• ×Joining the points of $xy=6$ with a straight line, or connecting its two branches through the origin. It is a curve, and no point has $x=0$. (Special Lines, xy = k and Inequalities)
• ×Changing the slope when translating. Moving $y=2x$ up 3 gives $y=2x+3$, not $y=5x$: a translation slides the line without tilting it. (Translating, Reflecting and Rotating Lines)
• ×Using $-a$ or $\frac{1}{a}$ for a 90° rotation. The rotated slope is the negative reciprocal, $-\frac{1}{a}$: for $y=4x$ it is $-\frac{1}{4}$. (Translating, Reflecting and Rotating Lines)
• ×Rotating a point with $(x,y)\to (y,x)$. That swap is a reflection in the line $y=x$; a 90° counterclockwise rotation is $(x,y)\to (-y,x)$. (Translating, Reflecting and Rotating Lines)
• ×Assuming the $x$-axis and $y$-axis reflections always give the same line. They agree for $y=ax$, but for $y=2x+3$ they give $y=-2x-3$ and $y=-2x+3$. (Translating, Reflecting and Rotating Lines)
• ×Applying $-\frac{1}{a}$ to $y=0$. You cannot divide by 0: a quarter turn of the $x$-axis gives the vertical line $x=0$. (Translating, Reflecting and Rotating Lines)

Practice test

24 questions from across the unit. Each one shows the lesson it comes from.

Endless drill

Fresh numbers every time, for the calculation questions in this unit.