Slope and y-intercept

In \(x=-4\), the x-coordinate is fixed, so the graph is a vertical line parallel to the y-axis. It cannot meet the y-axis, where \(x=0\). In contrast, \(y=-4\) is horizontal. Exam tip: \(x=a\) always represents a vertical line.

In the form \(y=mx+c\), the constant term \(c\) is the \(y\)-intercept. The first line has \(y\)-intercept \(-4\), while the second has \(k-1\). For equal intercepts, \(k-1=-4\), so \(k=-3\). If \(-2\) were used, the second intercept would be \(-3\), not \(-4\). Exam tip: set \(x=0\) to identify a \(y\)-intercept.

In the form \(y=mx+c\), the coefficient of \(x\) is the slope. The slopes of the first and second lines are \(m-1\) and \(2\), respectively. Equal slopes give \(m-1=2\), so \(m=3\). If 2 were chosen, the first slope would be 1, not equal to the second slope. Exam tip: identify the coefficient of \(x\) directly to find the slope.

In \(y=mx+c\), \(m\) is the slope and \(c\) is the y-intercept. In option A, \(m=-2\) is negative while \(c=5\) is positive. Option C has a negative slope too, but its intercept is negative. Exam tip: check the signs of the x-coefficient and constant separately.

The slope is \(\frac{5}{2}\), meaning that an increase of 2 units in \(x\) increases \(y\) by 5 units. Hence, \(\Delta y=\frac{5}{2}\Delta x\). Putting \(\Delta y=20\), we get \(20=\frac{5}{2}\Delta x\), so \(\Delta x=20\times\frac{2}{5}=8\). If \(x\) increased by 10, \(y\) would increase by 25, not 20. Exam tip: For a line \(y=mx+c\), use \(\Delta y=m\Delta x\) in change-based questions.

In y=px+q, p is the slope and q is the y-intercept. Since p=-4 is negative, y decreases as x increases, so the line decreases. Since q=9 is positive, the line cuts the y-axis above the origin. Option C is incorrect because a horizontal line must have slope 0. Exam tip: use the sign of p to identify whether the line rises or falls, and the sign of q to locate the y-intercept.

In \(x=-3\), the x-coordinate stays fixed, so the graph is a vertical line parallel to the y-axis. It cannot meet the y-axis, where \(x=0\). In contrast, \(x=0\) is the y-axis itself. Exam tip: a constant-x line has undefined slope.

Writing the given line as \(y=-\frac{3}{2}x+4\) gives slope \(-\frac{3}{2}\) and y-intercept 4. A perpendicular line has slope \(\frac{2}{3}\), so \(y=\frac{2}{3}x+4\), which is option C. Option B has intercept 4 but slope \(-\frac{2}{3}\). Exam tip: perpendicular slopes multiply to \(-1\).

A line in the form y=mx+c has y-intercept c. Here, 2n+1=11, so 2n=10 and n=5. Therefore, the slope is n-5=5-5=0. Option 1 cannot be the slope because substituting n=5 makes the coefficient of x zero. Exam tip: In y=mx+c, the coefficient of x is always the slope.

A line in the form \(y=mx+c\) has slope \(m\) and y-intercept \(c\). Here, the slope is \(r+4\). Thus, \(r+4=1\), giving \(r=-3\). Therefore, the y-intercept is \(r-2=-3-2=-5\). Note that \(-3\) is the value of \(r\), not the y-intercept. Exam tip: first find the parameter from the slope, then substitute it into the constant term.

An increase of \(10\) in \(y\) for an increase of \(2\) in \(x\) gives the slope \(m=\frac{10}{2}=5\). Since the line crosses the \(y\)-axis at \((0,-6)\), its \(y\)-intercept is \(b=-6\). Using \(y=mx+b\), the equation is \(y=5x-6\). The option \(y=10x-6\) has the correct intercept but an incorrect slope of \(10\). Exam tip: find the slope from the rate of change first, then identify the intercept by putting \(x=0\).

When \(x\) increases by 3 and \(y\) decreases by 12, the slope is \(m=\frac{-12}{3}=-4\). Since the line cuts the \(y\)-axis at \((0,9)\), its \(y\)-intercept is 9. Therefore, its equation is \(y=-4x+9\). Option A has a positive slope, but \(y\) is decreasing here. Exam tip: Use a negative sign for the slope whenever the dependent variable decreases.

In \(y=mx+c\), \(m=-\frac{2}{3}\) is negative, so the line falls from left to right. Since \(c=4\), its y-intercept is \((0,4)\). Exam tip: use \(m\) for direction and \(c\) for the y-intercept.

In the form y = mx + c, the y-intercept is the constant term c. Here, the y-intercept is t + 4. From t + 4 = 10, we get t = 6. Therefore, the slope is 2t - 3 = 2(6) - 3 = 9. Option 6 is the value of t, not the slope. Exam tip: set x = 0 to identify the y-intercept.

In the line (y=mx+c), c is the y-intercept because x=0 on the y-axis. The first line has y-intercept b, while the second has y-intercept 4. For both lines to meet the y-axis at the same point, b=4. The terms 3x and -2x represent slopes, so they do not determine the y-intercept. Exam tip: put x=0 to find a y-intercept.

In the form \(y=mx+c\), the coefficient of \(x\) is the slope. Since the two lines have equal slopes, \(a=4\). Substituting \(x=2\) in the first line gives \(y=4\times2-5=3\). Option 4 is the value of the slope, not the value of \(y\). Exam tip: first equate the slopes to find the unknown coefficient, then substitute the given value of \(x\).

In the form y=-7x+c, the y-intercept is c. Since the y-intercept is 3, c=3. Substituting x=-2 gives y=-7(-2)+3=14+3=17. Hence, 17 is correct. The value 14 comes only from -7(-2); the y-intercept 3 must also be added. Exam tip: When the y-intercept is given, use it as the constant term c.

From \(3x-6=0\), we get \(x=2\). A line with a constant \(x\)-value is vertical, so it is parallel to the \(y\)-axis and has an undefined slope. Exam tip: whenever \(x=c\), identify the slope as undefined.

The slope of the line is \(-(p+1)\). Since an increase of 3 in \(x\) causes a decrease of 15 in \(y\), the slope is \(\frac{-15}{3}=-5\). Therefore, \(-(p+1)=-5\), so \(p+1=5\) and \(p=4\). Option 5 would result from incorrectly taking \(p+1\) as the answer. Exam tip: for change-based questions, find the slope using \(\frac{\Delta y}{\Delta x}\).

In \(y=mx+c\), \(m\) is the slope and \(c\) is the y-intercept. For \(y=-2x\), \(m=-2\) and \(c=0\), so it passes through the origin. Exam tip: check \(c=0\) first for an origin-passing line.

Given \(m+c=10\) and \(c=4\), substitute the value of \(c\): \(m+4=10\). Hence, \(m=6\). In the linear equation \(y=mx+c\), \(m\) represents the slope, so the correct answer is 6. The value 4 is only \(c\), the y-intercept, not the slope. Exam tip: in \(y=mx+c\), the coefficient of \(x\), \(m\), is the slope.

In the form y=mx+c, m is the slope and c is the y-intercept. Given m=-3, substitute it into m+c=5: -3+c=5, so c=8. Therefore, the y-intercept is 8. The value -8 can result from a sign error; moving -3 to the other side means adding 3. Exam tip: isolate c to find the y-intercept.

Move \(8x\) to the other side: \(4y=-8x+28\). Dividing by 4 gives \(y=-2x+7\). Comparing this with \(y=mx+c\), we get \(m=-2\) and \(c=7\). In option A, the y-intercept is correct, but the negative sign of the slope is missing. Exam tip: For \(Ax+By=C\), divide the whole equation by the coefficient of \(y\) after isolating the \(y\)-term.

The line x = 4 is parallel to the y-axis, so it never meets the y-axis and has no y-intercept. In contrast, y = 4 meets it at (0, 4). Exam tip: x = constant denotes a vertical line.

A line in the form \(y=mx+c\) has slope \(m\). Here, the coefficient of \(x\) is \(k+4\), so \(k+4=9\). Therefore, \(k=9-4=5\). Option \(9\) is the slope, not the value of \(k\). Exam tip: In \(y=mx+c\), identify the coefficient of \(x\) as the slope.