Slope and Graphing Linear Functions: practice solutions, Grade 9 – download the PDF

1 Shrinking candle ★★★

1. Rate \(=\dfrac{9-18}{4-1}=\dfrac{-9}{3}=-3\) cm per hour.
2. The negative sign means the height decreases: the candle loses 3 cm every hour.
3. From hour 4 to hour 6 there are 2 more hours, so the candle loses \(2\times 3=6\) cm: \(9-6=3\) cm. After 6 hours the candle is 3 cm tall (about 1.2 inches).

2 Four slopes ★★★

1. \(m=\dfrac{11-3}{6-2}=\dfrac84=2\).
2. \(m=\dfrac{-12-4}{5-(-3)}=\dfrac{-16}{8}=-2\).
3. The \(x\)-values are equal, so the run is 0: the line is vertical and the slope is undefined.
4. \(m=\dfrac{0-0}{0-(-5)}=0\): the line is horizontal.

3 Read the slope and intercept ★★★

1. \(m=5\), \(b=-7\).
2. \(m=-\dfrac34\), \(b=2\).
3. Rewrite as \(y=-x+6\): \(m=-1\), \(b=6\).
4. Rewrite as \(y=\dfrac12x+0\): \(m=\dfrac12\), \(b=0\).

4 Write the equation ★★★

1. \(y=-2x+9\).
2. \(y=\dfrac13x-4\).
3. \(y=0x+5\), that is \(y=5\): a horizontal line.

5 Find the intercepts ★★★

1. If \(y=0\): \(2x=8\), \(x=4\). If \(x=0\): \(y=8\). Intercepts: \((4,\,0)\) and \((0,\,8)\).
2. If \(y=0\): \(x=9\). If \(x=0\): \(-3y=9\), so \(y=-3\). Intercepts: \((9,\,0)\) and \((0,\,-3)\).

6 Direct variation or not? ★★★

1. Yes, \(k=7\).
2. No: the line does not go through the origin (when \(x=0\), \(y=1\)).
3. Yes, \(k=-\dfrac13\).
4. No: this is not of the form \(y=kx\), since \(x\) is in the denominator.

7 Read a graph ★★★

1. \(A\) to \(B\): \(m=\dfrac{1-(-2)}{0-(-2)}=\dfrac32\). \(B\) to \(C\): \(m=\dfrac{4-1}{2-0}=\dfrac32\). Both agree, so the slope is \(\dfrac32\).
2. The line crosses the \(y\)-axis at \(B(0,\,1)\), so \(b=1\) and \(y=\dfrac32x+1\).

8 Into standard form ★★★

1. Multiply by 5: \(5y=2x-15\). Move the \(x\)-term: \(-2x+5y=-15\). Multiply by \(-1\) so that \(A\) is positive: \(2x-5y=15\). Slope \(=-\dfrac{2}{-5}=\dfrac25\). Check with \((0,\,-3)\): \(0+15=15\).
2. Multiply by 4: \(4y=-3x+24\), so \(3x+4y=24\). Slope \(=-\dfrac34\). Check with \((0,\,6)\): \(24=24\).

9 Slope and a point ★★★

\(y-(-2)=5(x-3)\), so \(y+2=5x-15\) and \(y=5x-17\).

Check: for \(x=3\), \(5\cdot 3-17=-2\). The line goes through the point.

10 Through two points ★★★

Slope: \(m=\dfrac{-2-6}{3-(-1)}=\dfrac{-8}{4}=-2\).

Point-slope with \((-1,\,6)\): \(y-6=-2(x+1)\), so \(y-6=-2x-2\) and \(y=-2x+4\).

Check with \((3,\,-2)\): \(-2\cdot 3+4=-2\). Correct.

11 Parallel, perpendicular, or neither? ★★★

1. Same slope \(3\), different intercepts: parallel.
2. \(3\times\left(-\dfrac13\right)=-1\): perpendicular.
3. The first slope is \(-\dfrac23\). Solving the second for \(y\): \(-2y=-3x+8\), \(y=\dfrac32x-4\), slope \(\dfrac32\). Product: \(-\dfrac23\times\dfrac32=-1\), so the lines are perpendicular.
4. Slopes \(2\) and \(-2\): not equal, and the product is \(-4\), not \(-1\). Neither.

12 Gym membership ★★★

1. \(C=18m+25\).
2. The slope 18 is the monthly price (dollars per month); the \(y\)-intercept 25 is the sign-up fee paid at the start (\(m=0\)).
3. \(C=18\times 8+25=144+25=169\) dollars.
4. \(18m+25=277\), so \(18m=252\) and \(m=14\). After 14 months.

13 Fuel economy ★★★

1. \(k=\dfrac{128}{4}=32\) miles per gallon, so \(d=32g\).
2. \(d=32\times 9=288\) miles.
3. \(32g=400\), so \(g=\dfrac{400}{32}=12.5\) gallons.

14 Graph with intercepts ★★★

1. If \(y=0\): \(4x=12\), \(x=3\). If \(x=0\): \(3y=12\), \(y=4\). Plot \((3,\,0)\) and \((0,\,4)\) and draw the line through them.
2. \(m=-\dfrac43\).
3. \(4\cdot 6+3\cdot(-4)=24-12=12\). Yes, the point lies on the line.

15 Find the missing coordinate ★★★

1. \(\dfrac{9-3}{5-k}=2\), so \(5-k=3\) and \(k=2\).
2. \(\dfrac{6-k}{-2-4}=-\dfrac32\), so \(6-k=-6\times\left(-\dfrac32\right)=9\) and \(k=-3\). Check: \(\dfrac{6-(-3)}{-6}=-\dfrac32\).

16 A perpendicular line ★★★

The given line has slope \(-\dfrac34\) (since \(4y=-3x+8\)). The perpendicular slope is the opposite reciprocal, \(\dfrac43\).

Point-slope: \(y-1=\dfrac43(x-6)\), so \(y=\dfrac43x-8+1=\dfrac43x-7\).

Multiply by 3: \(3y=4x-21\), so \(4x-3y=21\). Check with \((6,\,1)\): \(24-3=21\).

17 A parallel line ★★★

The slope is \(-\dfrac25\). Point-slope: \(y-2=-\dfrac25(x-5)\), so \(y=-\dfrac25x+2+2=-\dfrac25x+4\).

Multiply by 5: \(5y=-2x+20\), so \(2x+5y=20\). Check with \((5,\,2)\): \(10+10=20\).

18 A right triangle? ★★★

\(m_{AB}=\dfrac{4-2}{5-1}=\dfrac12\), \(m_{BC}=\dfrac{8-4}{3-5}=-2\), \(m_{AC}=\dfrac{8-2}{3-1}=3\).

\(m_{AB}\times m_{BC}=\dfrac12\times(-2)=-1\), so \(AB\perp BC\). The other products are \(\dfrac12\times 3=\dfrac32\) and \(-2\times 3=-6\), neither equal to \(-1\). The triangle has a right angle at \(B\) only.

19 Taxi fare ★★★

1. Rate \(=\dfrac{26-12.5}{10-4}=\dfrac{13.5}{6}=2.25\) dollars per mile. Starting fee: \(12.5-2.25\times 4=12.5-9=3.5\) dollars.
2. \(C=2.25m+3.5\). Check: \(2.25\times 10+3.5=26\).
3. \(2.25m+3.5\le 40\), so \(2.25m\le 36.5\) and \(m\le 16.2\ldots\). The longest whole-number ride is 16 miles (cost \(36+3.5=39.5\) dollars; a 17-mile ride costs 41.75 dollars).

20 Fahrenheit and Celsius ★★★

1. \(m=\dfrac{212-32}{100-0}=\dfrac{180}{100}=\dfrac95\). The intercept is \(b=32\), so \(F=\dfrac95C+32\).
2. \(F=\dfrac95\times 35+32=63+32=95^\circ\text{F}\). For \(68^\circ\text{F}\): \(68=\dfrac95C+32\), so \(\dfrac95C=36\) and \(C=20^\circ\text{C}\).
3. Set \(C=\dfrac95C+32\): \(-\dfrac45C=32\), so \(C=-40\). Both scales read \(-40\).
4. Multiply \(F=\dfrac95C+32\) by 5: \(5F=9C+160\), so \(9C-5F=-160\).

21 Where two lines meet ★★★

The slope of \(\ell_2\) is \(-\dfrac12\), so \(\ell_2\): \(y=-\dfrac12x+9\).

Set the two expressions equal: \(2x-1=-\dfrac12x+9\), so \(\dfrac52x=10\) and \(x=4\). Then \(y=2\cdot 4-1=7\).

The lines cross at \((4,\,7)\). Check in \(\ell_2\): \(-2+9=7\).

22 True or false? ★★★

1. False. It is linear, but it crosses the \(y\)-axis at 3, not at the origin.
2. True. If \(y=kx\) then \(k(2x)=2(kx)=2y\).
3. True. The slope is \(\dfrac{15-6}{5-2}=3\), and \(3\times 2=6\), \(3\times 5=15\), so \(y=3x\) fits both points and the line passes through the origin.
4. False. A vertical line has an undefined slope; a horizontal line has slope \(0\).