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Slope and Intercepts: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 The equation label

The line $y=2x+c-5$ has y-intercept $(0,-1)$ . Find $c$ .
1. Hint 1
  The y-intercept is the output when the input is zero.
2. Hint 2
  Match the whole constant term to the given intercept height.
Answer

$c=4$ .

Full solution

At $x=0$ , the equation gives
$y=c-5$
Thus the intercept information requires
$c-5=-1$ $c=4$
The resulting line is $y=2x-1$ , whose output at zero is indeed $-1$ .

Answer

$c=4$ .

Key idea

The entire constant term in slope-intercept form gives the intercept height.
Problem 1: Mark complete
Problem 2 A divided record

Find the slope and y-intercept of the line $4(y+1)=12-5x$ .
1. Hint 1
  Both pieces of information can be read once $y$ stands alone.
2. Hint 2
  Divide both sides by 4, then subtract the amount needed to isolate $y$ .
Answer

Slope $-\frac54$ , or $-1.25$ ; y-intercept $(0,2)$ .

Full solution

Dividing both sides by 4 gives
$y+1=3-\frac54x$
Subtracting 1 and ordering the terms gives
$y=-\frac54x+2$
The slope is $-5/4$ and the intercept is $(0,2)$ .
In the original equation, $x=0$ gives $4(y+1)=12$ , confirming $y=2$ .

Answer

Slope $-\frac54$ , or $-1.25$ ; y-intercept $(0,2)$ .

Key idea

Read the slope and constant after all operations needed to isolate the output.
Problem 2: Mark complete
Problem 3 One parameter

For which values of $b$ does $y=bx-b$ pass through the origin?
1. Hint 1
  A line through the origin must give output zero at input zero.
2. Hint 2
  At $x=0$ , inspect the constant term rather than the coefficient of $x$ .
Answer

$b=0$ only.

Full solution

At the origin the equation becomes
$0=-b$
Thus $b=0$ is required.
For $b=0$ , the equation is $y=0$ , which contains the origin.

Answer

$b=0$ only.

Key idea

Passing through the origin forces the constant term to be zero.
Problem 3: Mark complete
Problem 4 Half-unit grid

On the blank grid in the figure, graph $2y=3x-1$ by marking its y-intercept and one point reached by a slope step. Also give its x-intercept exactly.

A blank coordinate grid with gridlines every half unit.

Text description of this figure

A blank rectangular coordinate grid. The horizontal x-axis runs from negative 2 to 3 and the vertical y-axis from negative 4 to 4, and each axis has an arrowhead at both ends. One unit has the same length on both axes. Gridlines are drawn every half unit, with the whole-number gridlines slightly darker than the half-unit ones. Each whole-number tick is labeled: negative 2, negative 1, 1, 2 and 3 on the x-axis, negative 4 through 4 on the y-axis, and the origin is labeled 0. No points, lines or other markings are drawn in the plotting area.
1. Hint 1
  Expose the slope and intercept by isolating $y$ .
2. Hint 2
  A rise of 3 for a run of 2 reaches another point; the x-intercept is where the output is zero.
Answer

Line through $(0,-\frac12)$ and, for example, $(2,\frac52)$ (or one slope step back, $(-2,-\frac72)$ ); x-intercept $(\frac13,0)$ .

Full solution

Dividing by 2 gives
$y=\frac32x-\frac12$
The intercept is $(0,-1/2)$ .
A move of 2 right and 3 up reaches $(2,5/2)$ ; extend the line through those points.

At the x-intercept,
$0=3x-1$ $x=\frac13$
Thus the crossing is $(1/3,0)$ .
The point $(2,5/2)$ checks in the original equation since both sides are 5.

Answer

Line through $(0,-\frac12)$ and, for example, $(2,\frac52)$ (or one slope step back, $(-2,-\frac72)$ ); x-intercept $(\frac13,0)$ .

Key idea

The y-intercept and a slope step locate the line, while the other crossing requires solving.
Problem 4: Mark complete
Problem 5 A burning candle

The height $h$ cm of a candle $t$ hours after it is lit satisfies $3t+2h=18$ until it burns out. With $t$ on the horizontal axis and $h$ on the vertical axis, find the slope of this line, and both of its intercepts as ordered pairs $(t,h)$ . Say what each intercept means for the candle.
1. Hint 1
  Treat $t$ as the input and $h$ as the output; solving for $h$ exposes the slope and the vertical intercept.
2. Hint 2
  The vertical intercept comes from $t=0$ ; the horizontal intercept comes from setting the height to zero.
Answer

Slope $-1.5$ , or $-\frac32$ , cm per hour; vertical intercept $(0,9)$ : the candle is 9 cm tall when lit; horizontal intercept $(6,0)$ : it burns out 6 hours after it is lit.

Full solution

Subtracting $3t$ from both sides gives
$2h=18-3t$
Dividing both sides by 2 and ordering the terms gives
$h=-1.5t+9$
So the slope is $-1.5$ cm per hour: the candle gets 1.5 cm shorter each hour.

At $t=0$ the height is 9, so the vertical intercept is $(0,9)$ .
The candle is 9 cm tall at the moment it is lit.

Setting $h=0$ in the original equation gives $3t=18$ , so $t=6$ and the horizontal intercept is $(6,0)$ .
The height reaches zero after 6 hours, which is when the candle burns out.
Both points check in the original equation, since $3(0)+2(9)$ and $3(6)+2(0)$ both equal 18.

Answer

Slope $-1.5$ , or $-\frac32$ , cm per hour; vertical intercept $(0,9)$ : the candle is 9 cm tall when lit; horizontal intercept $(6,0)$ : it burns out 6 hours after it is lit.

Key idea

In a context, the vertical intercept is the output when the input is zero, and the horizontal intercept is the input at which the output reaches zero.
Problem 5: Mark complete
Problem 6 A family record

A family of lines has slope $a-1$ , and each member passes through $(1,2a+1)$ , where $a$ is real. Starting from point-slope form at that point, write the equation with $y$ alone and read off its y-intercept. Then find the member whose x-intercept is $(2,0)$ .
1. Hint 1
  Point-slope form needs one point and the slope, and here both are written in terms of $a$ .
2. Hint 2
  Distribute $a-1$ , then move the constant across so $y$ stands alone; an x-intercept at $(2,0)$ means $y=0$ when $x=2$ .
Answer

$y=(a-1)x+a+2$ , with y-intercept $(0,a+2)$ ; the member with x-intercept $(2,0)$ has $a=0$ , which is $y=-x+2$ .

Full solution

Point-slope form at $(1,2a+1)$ with slope $a-1$ reads
$y-(2a+1)=(a-1)(x-1)$
Distributing $a-1$ on the right gives
$y-(2a+1)=(a-1)x-a+1$
Adding $2a+1$ to both sides gives
$y=(a-1)x+a+2$
Setting $x=0$ leaves $y=a+2$ , so the y-intercept is $(0,a+2)$ .

The x-intercept is $(2,0)$ exactly when $y=0$ at $x=2$ .
Substituting gives
$2(a-1)+a+2=0$
The left side simplifies to $3a$ , so $3a=0$ and $a=0$ .

With $a=0$ the equation is $y=-x+2$ .
It passes through $(1,1)$ , which is the point $(1,2a+1)$ when $a=0$ , and through $(2,0)$ , as required.

Answer

$y=(a-1)x+a+2$ , with y-intercept $(0,a+2)$ ; the member with x-intercept $(2,0)$ has $a=0$ , which is $y=-x+2$ .

Key idea

Solving a point-slope equation for $y$ exposes the y-intercept, and setting $y=0$ then locates the x-intercept.
Problem 6: Mark complete
Problem 7 Three special lines

On the blank grid in the figure, draw the lines A: $y=-3$ , B: $x=4$ and C: $y=-2x$ . For each line, give its y-intercept and its x-intercept, or state that it has none.

A blank coordinate grid with gridlines every unit.

Text description of this figure

A blank rectangular coordinate grid. The horizontal x-axis runs from negative 4 to 5 and the vertical y-axis from negative 5 to 5, and each axis has an arrowhead at both ends. One unit has the same length on both axes, and gridlines are drawn every unit. Each whole-number tick is labeled: negative 4 through 5 on the x-axis, negative 5 through 5 on the y-axis, and the origin is labeled 0. No points, lines or other markings are drawn in the plotting area.
1. Hint 1
  A y-intercept needs a point with $x=0$ and an x-intercept needs a point with $y=0$ ; ask whether each line has one.
2. Hint 2
  A keeps the same height and B keeps the same input everywhere; for C, find a second point by stepping the slope from $(0,0)$ .
Answer

A: y-intercept $(0,-3)$ , no x-intercept; B: no y-intercept, x-intercept $(4,0)$ ; C: both intercepts are $(0,0)$ . Drawn: A horizontal through $(0,-3)$ , B vertical through $(4,0)$ , C through $(0,0)$ and $(1,-2)$ .

Full solution

A is $y=-3$ , a horizontal line.
Setting $x=0$ leaves $y=-3$ , so its y-intercept is $(0,-3)$ .
Setting $y=0$ would require $0=-3$ , which is false, so A has no x-intercept.
Draw A as the horizontal line through $(0,-3)$ .

B is $x=4$ , a vertical line.
Setting $y=0$ leaves $x=4$ , so its x-intercept is $(4,0)$ .
Setting $x=0$ would require $0=4$ , which is false, so B has no y-intercept.
Draw B as the vertical line through $(4,0)$ .

C is $y=-2x$ .
Setting $x=0$ gives $y=0$ , and setting $y=0$ gives $x=0$ , so both intercepts are the origin.
That single point does not fix the line, so step by the slope $-2$ : right 1 and down 2 reaches $(1,-2)$ .
Draw C through $(0,0)$ and $(1,-2)$ , and label the three lines A, B and C.

Answer

A: y-intercept $(0,-3)$ , no x-intercept; B: no y-intercept, x-intercept $(4,0)$ ; C: both intercepts are $(0,0)$ . Drawn: A horizontal through $(0,-3)$ , B vertical through $(4,0)$ , C through $(0,0)$ and $(1,-2)$ .

Key idea

A horizontal line off the x-axis has no x-intercept, a vertical line off the y-axis has no y-intercept, and a line through the origin with nonzero slope has both intercepts at $(0,0)$ .
Problem 7: Mark complete
Problem 8 Two constant settings

A student says that increasing $b$ by 4 in $y=2x+b$ also increases the x-intercept coordinate by 4. Is this correct? Justify your answer by finding the x-intercepts of the lines with $b=6$ and $b=10$ .
1. Hint 1
  The x-intercept coordinate must be found from output zero.
2. Hint 2
  For each value of $b$ , set $2x+b=0$ and solve for $x$ , then compare the two results.
Answer

No; the crossings are $(-3,0)$ and $(-5,0)$ , so the x-intercept coordinate decreases by 2.

Full solution

For $b=6$ , the crossing satisfies $0=2x+6$ , giving $x=-3$ .
For $b=10$ , it satisfies $0=2x+10$ , giving $x=-5$ .

The change is
$-5-(-3)=-2$
Thus it decreases by 2, not increases by 4.
In general here $x=-b/2$ , which agrees with the computed change.

Answer

No; the crossings are $(-3,0)$ and $(-5,0)$ , so the x-intercept coordinate decreases by 2.

Key idea

For $y=mx+b$ with $m\ne0$ , the x-intercept is at $x=-b/m$ , so a change in $b$ moves it by that change divided by $-m$ .
Problem 8: Mark complete
Problem 9 Opposite axis signs

A nonvertical line crosses the y-axis above the origin and crosses the x-axis to the left of the origin. A student claims its slope must be positive. Is the claim correct? Explain using the two crossings.
1. Hint 1
  Imagine traveling from the left crossing to the upper crossing.
2. Hint 2
  Track the signs of the rise and run in that direction.
Answer

Yes; its slope is positive.

Full solution

Write the crossings as $(a,0)$ and $(0,b)$ , where $a$ is negative and $b$ is positive.
Traveling from the first to the second gives
$m=\frac{b}{-a}$
The numerator is positive and the denominator is also positive, so the ratio is positive.

The nonzero run makes this division valid.
The line rises as it moves right from one crossing to the other, checking the sign.

Answer

Yes; its slope is positive.

Key idea

The positions of two axis crossings can determine the sign of a slope.
Problem 9: Mark complete
Problem 10 A total of heights

Two points on a line have coordinates $(0,b)$ and $(1,c)$ . You are told only that $b+c=6$ . Is that enough to determine the slope? Justify your answer.
1. Hint 1
  Write the slope between $(0,b)$ and $(1,c)$ in terms of $b$ and $c$ .
2. Hint 2
  Try two different pairs of heights with the same sum and calculate their slopes.
Answer

No; for example, $y=4x+1$ and $y=2x+2$ have slopes $4$ and $2$ .

Full solution

The run between the points is 1, so their slope is
$m=c-b$
Choose $b=1$ and $c=5$ .
Their sum is 6 and the line is $y=4x+1$ .
Choose instead $b=2$ and $c=4$ .
Their sum is also 6 and the line is $y=2x+2$ .

The slopes differ, so the given sum does not determine a slope.
Each proposed line gives its chosen heights at inputs 0 and 1.

Answer

No; for example, $y=4x+1$ and $y=2x+2$ have slopes $4$ and $2$ .

Key idea

A fixed sum of heights does not fix the difference that determines slope.
Problem 10: Mark complete

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