Free math worksheets

- Where on the coordinate plane is the function value equal to \(0\)? - What happens when a line crosses the x-axis? - What is the y-coordinate of every point on the x-axis?

1. The graph crosses the x-axis where \(f(x) = 0\), so solve \(-2x + 6 = 0\). 2. Then \(-2x = -6\), so \(x = 3\). 3. Every point on the x-axis has y-coordinate \(0\), so the x-intercept is \((3, 0)\). 4. The x-coordinate of this intercept, \(3\), is exactly the solution of the equation \(-2x + 6 = 0\).

The graph crosses the x-axis at \((3, 0)\). The x-coordinate \(3\) is the solution of the original equation.

- Locate the point where the line crosses the y-axis. - Choose another lattice point on the line and compare rise with run. - Use the graph's direction to check the sign of your slope.

1. The graph crosses the y-axis at \((0,-3)\), so the y-intercept is \(-3\). 2. From \((0,-3)\) to \((1,-1)\), the graph rises \(2\) units while moving right \(1\) unit, so the slope is \(2\).

Slope: \(2\); y-intercept: \(-3\)

- A zero is an input that produces an output of \(0\). - The y-intercept occurs when \(x=0\). - A point must satisfy the function equation to lie on the graph.

1. \(f(-2)=1.5(-2)+3=-3+3=0\), so statement a is true. 2. \(f(0)=1.5(0)+3=3\), so the y-intercept is \((0, 3)\). Statement b is true. 3. \(f(2)=1.5(2)+3=6\), not \(0\). Although \((2, 0)\) lies on the x-axis, it does not lie on the graph of \(f\). Statement c is false.

a) True, because \(f(-2)=0\). b) True, because \(f(0)=3\), so the y-intercept is \((0,3)\). c) False, because \(f(2)=6\), so \((2,0)\) is not on the graph.

- Find the entries where the output is \(0\). - The y-intercept has input \(x=0\). - Match the input \(4\) with its output.

1. Zeros occur where \(g(x)=0\). The table shows \(g(-1)=0\) and \(g(2)=0\), so the zeros are \(-1\) and \(2\). 2. The y-intercept occurs when \(x=0\). Since \(g(0)=-4\), the y-intercept is \((0, -4)\). 3. The table shows \(g(4)=5\).

1) \(-1\) and \(2\) 2) \((0, -4)\) 3) \(5\)

- What is the function value at a zero? - Set each function equal to \(0\) and solve the resulting linear equation for \(x\).

1. A zero occurs where the function value is \(0\). 2. For \(g_1\), solve \(0 = 3.2x + 8\). Then \(-8 = 3.2x\), so \(x = -2.5\). 3. For \(g_2\), solve \(0 = \frac{3}{5}x - 9\). Then \(9 = \frac{3}{5}x\), so \(x = 9 \cdot \frac{5}{3} = 15\).

a) \(x = -2.5\) b) \(x = 15\)

- What coordinate is always \(0\) at an axis intercept? - Read each y-intercept directly from the constant term. - Set each function equal to \(0\) to find its x-intercept. - Compare the two points where the graphs cross the x-axis.

1. For \(f\), the y-intercept is \((0, -1.5)\). To find the x-intercept, solve \(0 = \frac{3}{4}x - 1.5\), giving \(x = 2\). Thus the x-intercept is \((2, 0)\). 2. For \(g\), the y-intercept is \((0, 4)\). To find the x-intercept, solve \(0 = -2x + 4\), giving \(x = 2\). Thus the x-intercept is \((2, 0)\). 3. Both graphs have the same x-intercept, \((2, 0)\).

\(f\): y-intercept \((0, -1.5)\), x-intercept \((2, 0)\) \(g\): y-intercept \((0, 4)\), x-intercept \((2, 0)\) Both graphs have the same x-intercept.

- Evaluate the function at each point's x-coordinate. - Compare the point's y-coordinate with the function value. - Equality means the point lies on the graph.

1. \(k(2)=1.5(2)-3=0\). The point's y-coordinate equals the function value, so \(A\) lies on the graph. 2. \(k(4)=1.5(4)-3=3\). Since \(5>3\), \(B\) lies above the graph. 3. \(k(-2)=1.5(-2)-3=-6\). Since \(-7<-6\), \(C\) lies below the graph.

a) on the graph b) above the graph c) below the graph

- What y-value does every point on the x-axis have? - Evaluate \(g\) at the x-coordinate of \(N_f\). - Compare the two y-values at that same x-coordinate.

1. Set \(f(x) = 0\): \(1.2x - 3 = 0\), so \(1.2x = 3\) and \(x = 2.5\). Thus \(N_f = (2.5, 0)\). 2. Evaluate \(g\) at the same x-coordinate: \(g(2.5) = 0.5 \cdot 2.5 + 1 = 2.25\). 3. Since the point \(N_f\) has y-coordinate \(0\) and \(0 < 2.25\), \(N_f\) lies below the graph of \(g\).

a) Zero: \(x = 2.5\); x-intercept: \(N_f = (2.5, 0)\) b) \(N_f\) lies below the graph of \(g\).

- What boundaries enclose the region in the first quadrant? - Find where the graph crosses each coordinate axis. - What geometric shape is formed by the line and the axes?

1. The y-intercept is \((0, 4)\). 2. For the x-intercept, solve \(0 = -0.8x + 4\), giving \(x = 5\). Thus the x-intercept is \((5, 0)\). 3. The enclosed region is a right triangle with leg lengths \(4\) and \(5\). Its area is \(\frac{1}{2} \cdot 5 \cdot 4 = 10\) square units.

\(10\) square units

- What does the constant term tell you about the y-intercept? - To find an x-intercept, set the function equal to \(0\). - Compare the slopes. What happens when only the sign of the x-term changes?

1. For \(f(x) = 2x - 4\), the y-intercept is \((0, -4)\). Solving \(2x - 4 = 0\) gives the x-intercept \((2, 0)\). 2. For \(g(x) = 2x + 4\), the y-intercept is \((0, 4)\), and solving \(2x + 4 = 0\) gives the x-intercept \((-2, 0)\). Compared with \(f\), the y-intercept moves from \(-4\) to \(4\), and the x-intercept moves from \(2\) to \(-2\). 3. For \(h(x) = -2x - 4\), the y-intercept remains \((0, -4)\), and solving \(-2x - 4 = 0\) gives the x-intercept \((-2, 0)\). Since \(h(x) = f(-x)\), its graph is the reflection of the graph of \(f\) across the y-axis.

a) The y-intercept changes from \((0, -4)\) to \((0, 4)\), and the x-intercept changes from \((2, 0)\) to \((-2, 0)\). b) The y-intercept stays at \((0, -4)\), and the x-intercept changes from \((2, 0)\) to \((-2, 0)\).

- Set each function equal to \(0\) to find its zero. - Two graphs intersect where they have the same output for the same input. - Compare the point's y-coordinate with \(g(2)\).

1. Solve \(1.5x-3=0\) to get \(x=2\). Solve \(-0.5x+5=0\) to get \(x=10\). 2. \(f(4)=1.5(4)-3=3\) and \(g(4)=-0.5(4)+5=3\). Therefore, \(P(4, 3)\) is the intersection point of the two graphs. 3. \(g(2)=-0.5(2)+5=4\). Since the point's y-coordinate \(1\) is less than \(4\), \(U\) lies below the graph of \(g\).

a) Zero of \(f\): \(2\); zero of \(g\): \(10\) b) \(P(4, 3)\) is the intersection point. c) \(U(2, 1)\) lies below the graph of \(g\).

- Which x-values make a fractional slope produce integer changes in \(y\)? - The y-intercept is often a convenient first point. - How can you rewrite the decimal slope as a fraction?

1. For \(f(x) = \frac{3}{5}x - 1\), choose x-values that are multiples of \(5\). Then \(f(0) = -1\) and \(f(5) = 2\), giving \((0, -1)\) and \((5, 2)\). 2. For \(g(x) = -1.2x + 2\), write the slope as \(-1.2 = -\frac{6}{5}\). Choosing x-values that are multiples of \(5\) gives \(g(0) = 2\) and \(g(5) = -4\), so two convenient points are \((0, 2)\) and \((5, -4)\).

a) One possible pair is \((0, -1)\) and \((5, 2)\). b) One possible pair is \((0, 2)\) and \((5, -4)\).

- Think about where each graph crosses the coordinate axes. - Use the intercepts and slope to reason about where each graph lies. - What does a positive or negative slope tell you about the direction of a line? - What is special about a line whose y-intercept is \(0\)?

1. For \(f(x) = 1.5x - 3\), the y-intercept is \((0, -3)\) and the x-intercept is \((2, 0)\). For negative \(x\), both coordinates are negative, so the line passes through Quadrant III. For \(0 < x < 2\), it is in Quadrant IV, and for \(x > 2\), it is in Quadrant I. Thus, it passes through Quadrants I, III, and IV. 2. The graph of \(g(x) = -x\) passes through the origin with negative slope, so it passes through Quadrants II and IV. 3. The graph of \(h(x) = 4\) is a horizontal line above the x-axis, so it passes through Quadrants I and II.

1) Quadrants I, III, and IV 2) Quadrants II and IV 3) Quadrants I and II

- What function value corresponds to a zero? - Set each function equal to \(0\) and solve for \(x\). - For part d, simplify the expression before solving if that makes the equation easier to work with.

1. For part a, set \(6x + 18 = 0\). Then \(6x = -18\), so \(x = -3\). 2. For part b, set \(\frac{3}{4}x - 6 = 0\). Then \(\frac{3}{4}x = 6\), so \(x = 6 \cdot \frac{4}{3} = 8\). 3. For part c, set \(2.5 - 0.5x = 0\). Then \(0.5x = 2.5\), so \(x = 5\). 4. For part d, set \(4(x - 2) + 12 = 0\). Simplifying gives \(4x + 4 = 0\), so \(x = -1\).

a) \(x = -3\) b) \(x = 8\) c) \(x = 5\) d) \(x = -1\)

- What equation describes an x-intercept? - Does a vertical line pass the vertical line test? - What is the slope of a horizontal line, and when does such a line meet the x-axis?

1. For part a, an x-intercept satisfies \(mx + b = 0\). If \(m \neq 0\), this has exactly one solution, \(x = -\frac{b}{m}\). The statement is true. 2. For part b, a vertical line has equation \(x = c\). It cannot be written as \(y = mx + b\) and is not the graph of a function of \(x\). The statement is true. 3. For part c, a horizontal line has equation \(y = b\). If it is not the x-axis, then \(b \neq 0\), so the output is never \(0\). The statement is true.

a) True. If \(m\ne0\), solving \(mx+b=0\) gives the unique x-intercept \(x=-\frac{b}{m}\). b) True. A vertical line \(x=c\) cannot be written as \(y=mx+b\). c) True. A horizontal line \(y=b\) with \(b\ne0\) never reaches \(y=0\), so it has no zero.

- Where does the graph cross the y-axis? - Use the slope to find a second point from the y-intercept. - Which point on the line has y-coordinate \(1\)?

1. The y-intercept is \((0, 4)\) and the slope is \(-1.5 = -\frac{3}{2}\). 2. From \((0, 4)\), moving 2 units right and 3 units down gives another point, \((2, 1)\). 3. The point \((2, 1)\) shows that \(f(x) = 1\) when \(x = 2\). Equivalently, solving \(-1.5x + 4 = 1\) gives \(x = 2\).

The y-intercept is \((0, 4)\), and another point is \((2, 1)\). Therefore, \(f(x) = 1\) when \(x = 2\).

- Set the function rule equal to the target output. - What happens to every output when a graph shifts up 2 units? - Compare where the original and shifted functions reach the same output.

1. Solve \(\frac{1}{2}x + 1 = 3\). Then \(\frac{1}{2}x = 2\), so \(x = 4\). 2. Shifting \(f\) up 2 units gives \(g(x) = \frac{1}{2}x + 3\). Solving \(g(x) = 3\) gives \(\frac{1}{2}x + 3 = 3\), so \(x = 0\). 3. The upward shift raises every point while keeping the same slope. Therefore, the same target height \(y = 3\) is reached farther to the left, at a smaller x-value.

a) \(x = 4\) b) \(x = 0\) c) The upward shift makes the graph reach \(y = 3\) at a smaller x-value.

- You may choose a value for one variable and solve for the other. - Rewrite the equation in slope-intercept form to see how \(y\) depends on \(x\). - Try a negative x-value and check the resulting y-value.

1. If \(x = 0\), then \(-2y = 6\), so \(y = -3\). This gives \((0, -3)\). 2. If \(y = 0\), then \(3x = 6\), so \(x = 2\). This gives \((2, 0)\). 3. If \(x = 4\), then \(12 - 2y = 6\), so \(y = 3\). This gives \((4, 3)\). 4. Solving for \(y\) gives \(y = 1.5x - 3\). Choosing \(x = -2\) gives \(y = -6\), so \((-2, -6)\) is a solution with both coordinates negative.

a) One possible set is \((0, -3)\), \((2, 0)\), and \((4, 3)\). b) Yes. For example, \((-2, -6)\) satisfies the equation.

- Which coordinate stays constant on a vertical line? - What do all points on a horizontal line have in common? - Which axis is a line of the form \(x = \text{constant}\) or \(y = \text{constant}\) parallel to? - How can you describe each line's position relative to the coordinate axes?

1. A vertical line has the same x-coordinate at every point. Since it passes through \(A(3, -2)\), its equation is \(x = 3\). It is parallel to the y-axis. 2. A horizontal line has the same y-coordinate at every point. Since it passes through \(A\), its equation is \(y = -2\). It is parallel to the x-axis. 3. For example, \((3, 0)\) and \((3, 5)\) lie on \(g\), while \((0, -2)\) and \((1, -2)\) lie on \(h\).

a) \(x = 3\), parallel to the y-axis b) \(y = -2\), parallel to the x-axis c) One possible set is \((3, 0)\), \((3, 5)\) on \(g\), and \((0, -2)\), \((1, -2)\) on \(h\).

- What function value corresponds to a zero? - Solve each equation for \(x\). - Compare the two x-values on a number line.

1. For \(f\), solve \(2x - 5 = 0\). Then \(2x = 5\), so \(x = 2.5\). 2. For \(g\), solve \(-x + 4 = 0\), so \(x = 4\). 3. Since \(4 > 2.5\), the zero of \(g\) lies farther to the right on the x-axis.

The zero of \(f\) is \(x = 2.5\), and the zero of \(g\) is \(x = 4\). The zero of \(g\) lies farther to the right.

- Substitute the given x-values into both functions. - Compare each value of \(g(x)\) with the corresponding value of \(f(x)\). - Decide whether the outputs get larger or smaller and what that means for the graph. - Think about how multiplying a function's outputs by a number between \(0\) and \(1\) changes its graph.

1. Evaluate each function. For \(x = 4\), \(f(4) = 4\) and \(g(4) = \frac{1}{4} \cdot 4 = 1\). For \(x = 8\), \(f(8) = 8\) and \(g(8) = \frac{1}{4} \cdot 8 = 2\). 2. Since \(g(x) = \frac{1}{4}x = \frac{1}{4}f(x)\), for \(x \neq 0\) the ratio is \(\frac{g(x)}{f(x)} = \frac{1}{4}\). 3. Every output of \(g\) is one-fourth the corresponding output of \(f\), so the graph of \(g\) is a vertical compression of the graph of \(f\) by a factor of \(\frac{1}{4}\).

1. \(f(4) = 4\), \(g(4) = 1\); \(f(8) = 8\), \(g(8) = 2\). 2. \(\frac{g(x)}{f(x)} = \frac{1}{4}\) for \(x \neq 0\), equivalently \(g(x) = \frac{1}{4}f(x)\). 3. The graph of \(g\) is a vertical compression of the graph of \(f\) by a factor of \(\frac{1}{4}\).

- What changes about a point's y-coordinate when the point is reflected across the x-axis? - Try reflecting an intercept of the original line. - How can you represent changing every output to its opposite using the function rule?

1. A reflection across the x-axis keeps the x-coordinate and changes the sign of the y-coordinate, so \((x, y)\) maps to \((x, -y)\). 2. Therefore, every output of the original function is multiplied by \(-1\): \(h(x) = -(3x - 6)\). 3. Simplify: \(h(x) = -3x + 6\). 4. For example, \((0, -6)\) lies on \(g\), and its reflection \((0, 6)\) satisfies \(h(0) = 6\).

a) \(h(x) = -3x + 6\) b) A reflection across the x-axis maps \((x, y)\) to \((x, -y)\). Thus the new output is the opposite of the old output: \(h(x) = -g(x) = -(3x - 6) = -3x + 6\).

- Simplify each equation and notice which variable remains. - Think about how many y-values a fixed x-value would have on each line. - Relate the line's orientation to whether its rise/run ratio can be formed.

1. The first equation gives \(6y=18\), so \(g_1:y=3\). The second gives \(5x=-20\), so \(g_2:x=-4\). 2. The line \(g_1\) is the graph of a constant function because every x-value corresponds to the single y-value \(3\). The line \(g_2\) is not the graph of \(y=f(x)\) because \(x=-4\) corresponds to infinitely many y-values. 3. The horizontal line \(g_1\) has slope \(0\). The vertical line \(g_2\) has undefined slope. 4. The line \(g_1\) is horizontal and parallel to the x-axis. The line \(g_2\) is vertical and parallel to the y-axis.

1) \(g_1:y=3\); \(g_2:x=-4\) 2) \(g_1\) is a function; \(g_2\) is not. 3) \(g_1\) has slope \(0\); \(g_2\) has undefined slope. 4) \(g_1\) is horizontal and parallel to the x-axis; \(g_2\) is vertical and parallel to the y-axis.

- Read both intercepts for each line. - Use \(m=\frac{y_2-y_1}{x_2-x_1}\), then substitute the y-intercept into \(y=mx+b\).

1. Line \(f\) has y-intercept \((0, 4)\) and x-intercept \((4, 0)\). Its slope is \(m=\frac{0-4}{4-0}=-1\), so its equation is \(y=-x+4\). 2. Line \(g\) has y-intercept \((0, -2)\) and x-intercept \((1, 0)\). Its slope is \(m=\frac{0-(-2)}{1-0}=2\), so its equation is \(y=2x-2\).

Line \(f\): \(y=-x+4\); line \(g\): \(y=2x-2\)

- Read two exact points from the graph, such as the intercepts. - Use those points to determine the slope and a slope-intercept equation. - Rearrange the equation so the x- and y-terms are on the same side.

1. The graph crosses the y-axis at \((0,6)\) and the x-axis at \((3,0)\). 2. The slope is \(m=\frac{0-6}{3-0}=-2\), so the slope-intercept form is \(y=-2x+6\). 3. Move the x-term to the left: \(2x+y=6\).

\(2x+y=6\)

- First check whether each line has a special position: horizontal, vertical, or through the origin. - For a slanted line, use a slope triangle. - Pay attention to where each line crosses the axes.

1. Line \(f\) passes through the origin and \((2, 3)\). Its slope is \(\frac{3}{2}\), so \(f\) has equation \(y = \frac{3}{2}x\). 2. Line \(g\) is horizontal and crosses the y-axis at \(-2\), so its equation is \(y = -2\). 3. Line \(h\) is vertical and crosses the x-axis at \(-4\), so its equation is \(x = -4\).

\(f: y = \frac{3}{2}x\), \(g: y = -2\), and \(h: x = -4\)

- For a nonvertical line, first identify the y-intercept. - Use two grid points to calculate rise over run. - A vertical line is written using its constant x-coordinate.

1. Line \(p\) has y-intercept \(-2\). It rises \(2\) units for every \(3\) units to the right, so its slope is \(\frac{2}{3}\). Thus \(p: y = \frac{2}{3}x - 2\). 2. Line \(q\) has y-intercept \(1\). It falls \(1\) unit for every \(1\) unit to the right, so its slope is \(-1\). Thus \(q: y = -x + 1\). 3. Line \(r\) is vertical and passes through all points with x-coordinate \(-1\). Thus \(r: x = -1\).

\(p: y = \frac{2}{3}x - 2\); \(q: y = -x + 1\); \(r: x = -1\)

- Separate the equation into the two graph features given by slope-intercept form. - First eliminate graphs that rise in the wrong direction. - Then compare where the remaining graphs cross the y-axis.

1. The equation \(y=-x+2\) has slope \(-1\), so its graph must decrease from left to right. 2. Its y-intercept is \(2\), so it must cross the y-axis at \((0,2)\). 3. Graph b) has both features, so graph b) represents the equation.

Graph b). It has slope \(-1\) and y-intercept \(2\).

- Use the table row with input zero to identify one graph feature immediately. - Compare how much the output changes when the input increases by a known amount. - Check both features against each candidate graph; one correct feature is not enough.

1. The table shows a y-intercept of \(3\) because \(y=3\) when \(x=0\). 2. From \((0,3)\) to \((2,7)\), the slope is \(\frac{7-3}{2-0}=2\). 3. Graph c) has y-intercept \(3\) and slope \(2\), so it represents the function. The other graphs change either the intercept, the slope magnitude, or the slope sign.

c). The table gives y-intercept \(3\) and slope \(\frac{7-3}{2-0}=2\), and graph c) has both features.

- Use the two given points to determine the direction and steepness the correct graph must have. - Check where a candidate crosses the y-axis as well as whether it passes through the given points. - Eliminate graphs that get only one of those features right.

1. The slope through \(P\) and \(Q\) is \(\frac{5-1}{2-(-2)}=1\). 2. Using \(P(-2,1)\) in \(y=x+b\) gives \(1=-2+b\), so \(b=3\). 3. The line is \(y=x+3\), which is graph d). It has positive slope \(1\) and y-intercept \(3\).

d). The two points give slope \(1\), and \(1=-2+b\) gives y-intercept \(3\), so the line is \(y=x+3\).

- Think of a graph that never meets the x-axis. - Apply the function rule at \(x=0\). - Interpret what the coordinate \(y=0\) means.

1. Statement a is false. For example, the constant function \(f(x)=2\) never equals \(0\). 2. Statement b is false. A y-axis intersection has input \(x=0\), and a function can assign at most one output to that input. 3. Statement c is true. If \((4, 0)\) lies on the graph, then \(f(4)=0\), which means \(4\) is a zero.

a) False; for example, \(f(x)=2\) has no zero. b) False; a function can have at most one output at \(x=0\). c) True; \((4, 0)\) means \(f(4)=0\).

- What coordinate is \(0\) at each axis intercept? - The axes form the perpendicular legs of the triangle. - To test a point, evaluate the function at its x-coordinate and compare the result with its y-coordinate.

1. The y-intercept occurs at \(x = 0\), so \(f(0) = -3\). The y-intercept is \((0, -3)\). 2. For the x-intercept, solve \(0 = \frac{3}{4}x - 3\). This gives \(x = 4\), so the x-intercept is \((4, 0)\). 3. The coordinate axes form a right triangle with leg lengths \(4\) and \(3\). Its area is \(\frac{1}{2} \cdot 4 \cdot 3 = 6\) square units. 4. \(f(2) = \frac{3}{4} \cdot 2 - 3 = -1.5\). This matches the y-coordinate of \(Q\), so \(Q\) lies on the graph.

a) x-intercept \((4, 0)\); y-intercept \((0, -3)\) b) \(6\) square units c) Yes, \(Q(2, -1.5)\) lies on the graph.

- What is special about the y-intercept of a line through the origin? - How many zeros does a function have if its graph is the x-axis? - To find a zero algebraically, set the function equal to \(0\).

1. Since \(g_1\) passes through the origin, its y-intercept is \(0\). Using \(A(4, 2)\), \(2 = 4m\), so \(m = 0.5\). Thus \(g_1(x) = 0.5x\). Solving \(0.5x = 0\) gives \(x = 0\), so \(g_1\) has exactly one zero. 2. The equation is \(g_2(x) = -2x + 4\). Solving \(-2x + 4 = 0\) gives \(x = 2\), so \(g_2\) has exactly one zero. 3. A constant function has the form \(g_3(x) = c\). Since its graph passes through \((1, 0)\), \(c = 0\). Thus \(g_3(x) = 0\). Every real number is a zero, so \(g_3\) has infinitely many zeros.

1. \(g_1(x) = 0.5x\); one zero at \(x = 0\). 2. \(g_2(x) = -2x + 4\); one zero at \(x = 2\). 3. \(g_3(x) = 0\); infinitely many zeros.

- First find where the graph of \(f\) crosses the x-axis. - What must be true about \(g\) at that same x-value? - Use that condition to write an equation involving \(a\). - Solve the resulting equation for the parameter.

1. Find the zero of \(f\): \(2x - 6 = 0\), so \(2x = 6\) and \(x = 3\). 2. For the graphs to have the same x-intercept, \(g\) must also have a zero at \(x = 3\). 3. Set \(g(3) = 0\): \(3a + 3 = 0\). 4. Solving gives \(3a = -3\), so \(a = -1\).

\(a = -1\)

- What output value corresponds to an x-intercept, and what input value corresponds to a y-intercept? - Compare the two constant terms to determine the vertical shift. - What happens to a graph when \(x\) is replaced by \(-x\)?

1. The y-intercept is found at \(x = 0\), so it is \((0, -2)\). For the x-intercept, solve \(0 = \frac{2}{5}x - 2\), giving \(x = 5\). Thus, the x-intercept is \((5, 0)\). 2. Changing the constant term from \(-2\) to \(2\) increases every output by \(4\), so the graph shifts up 4 units. 3. The changed function is \(-\frac{2}{5}x - 2 = f(-x)\), so its graph is the reflection of the original graph across the y-axis.

1. x-intercept: \((5, 0)\); y-intercept: \((0, -2)\) 2. The graph shifts up 4 units. 3. The graph is reflected across the y-axis.

- Imagine how the line changes when you change \(m\) or \(b\). - If \(b\) is positive, where does the line cross the y-axis relative to the origin? - For an increasing line, when does the x-intercept lie to the left of the origin? - A straight line has no bends. How does that limit the quadrants it can pass through?

1. To pass from Quadrant III through Quadrant II and then into Quadrant I, the line must have positive slope, so \(m > 0\). It must also cross the y-axis above the origin, so \(b > 0\). Its x-intercept is \(-\frac{b}{m} < 0\), which places the three portions of the line in Quadrants III, II, and I. Thus, \(m > 0\) and \(b > 0\). 2. A nonvertical line can cross each coordinate axis at most once. Those crossings divide the line into at most three portions lying in open quadrants. Because a straight line cannot bend and cross into a fourth quadrant, the graph of a linear function cannot pass through all four quadrants.

a) \(m > 0\) and \(b > 0\) b) A straight line can pass through at most three quadrants.

- Substitute the known slope and point into slope-intercept form. - An x-intercept has output \(0\). - For the graph choice, check both intercepts and the direction/steepness of the line.

1. Substitute the slope and point into slope-intercept form: \(1=-0.5\cdot4+b\). Thus, \(b=3\), so \(k(x)=-0.5x+3\). 2. Set \(k(x)=0\): \(0=-0.5x+3\), so \(x=6\). The x-intercept is \((6,0)\). 3. Graph D has y-intercept \((0,3)\), slope \(-0.5\), and x-intercept \((6,0)\), so it agrees with the calculation.

1) \(k(x)=-0.5x+3\) 2) \((6,0)\) 3) D

- If an x-value is a zero, what is the function value there? - How does adding \(5\) to the entire function change its equation?

1. Since \(x = 2.5\) is a zero of \(f\), set \(f(2.5) = 0\): \(2.5a - 10 = 0\). Then \(2.5a = 10\), so \(a = 4\). 2. With \(a = 4\), \(f(x) = 4x - 10\), so \(g(x) = 4x - 10 + 5 = 4x - 5\). 3. Set \(g(x) = 0\): \(4x - 5 = 0\). Then \(4x = 5\), so \(x = 1.25\).

1. \(a = 4\) 2. The zero of \(g\) is \(x = 1.25\).

- What happens to an equation when you subtract the same value from both sides? - How does subtracting a constant from a function affect its graph? - Do the two graphical interpretations produce the same x-value?

1. Start with \(x + 2 = 5\). 2. Subtract \(5\) from both sides: \(x + 2 - 5 = 0\), which simplifies to \(x - 3 = 0\). 3. The expression on the left is \(h(x) = x - 3\), so solving the equation is the same as finding the zero of \(h\). 4. Solving \(x - 3 = 0\) gives \(x = 3\). 5. Graphically, subtracting \(5\) shifts both original graphs down by \(5\) units. Their intersection keeps the same x-coordinate, and the horizontal line \(y = 5\) becomes the x-axis.

Subtract \(5\) from both sides to rewrite the equation as \(x - 3 = 0\). This makes the solution the zero of \(h(x) = x - 3\). The solution is \(x = 3\).

- When do two distinct lines with finite slopes fail to intersect? - Consider horizontal, vertical, and slanted lines separately. - Simplify \(y = 0x + 0\) before thinking about its zeros.

1. Two distinct nonvertical lines have no common point when they are parallel: \(m_g = m_h\) and \(b_g \neq b_h\). 2. No. A nonhorizontal, nonvertical line intersects both axes. A horizontal line \(y = c\) intersects the y-axis, and a vertical line \(x = c\) intersects the x-axis. Thus, every line intersects at least one coordinate axis. 3. The equation \(y = 0x + 0\) simplifies to \(y = 0\), which is the x-axis. Every real \(x\) gives output \(0\), so the function has infinitely many zeros.

a) \(m_g = m_h\) and \(b_g \neq b_h\) b) No. Every line intersects at least one coordinate axis. c) It is the x-axis, and the function has infinitely many zeros.

- A point on the graph must make the function equation true. - What x-value identifies a point on the y-axis? - What y-value identifies a point on the x-axis?

1. Substitute \((4, -2)\) into the function: \(-2 = 0.5 \cdot 4 + b\). Thus \(b = -4\), so \(f(x) = 0.5x - 4\). 2. The y-intercept occurs at \(x = 0\), giving \((0, -4)\). 3. For the x-intercept, set \(f(x) = 0\): \(0 = 0.5x - 4\), so \(x = 8\). The x-intercept is \((8, 0)\).

\(b = -4\). The y-intercept is \((0, -4)\), and the x-intercept is \((8, 0)\).

- Find the slope from the two given points. - Which geometric shape is formed by the line and the two coordinate axes in the first quadrant? - Use the intercepts as the base and height of that shape.

1. Find the slope: \(m = \frac{-3 - 6}{4 - (-2)} = \frac{-9}{6} = -1.5\). 2. Substitute \((-2, 6)\) into \(y = -1.5x + b\): \(6 = -1.5 \cdot (-2) + b\), so \(b = 3\). Thus \(f(x) = -1.5x + 3\). 3. The y-intercept is \((0, 3)\). For the x-intercept, solve \(0 = -1.5x + 3\), giving \(x = 2\). 4. The triangle has base \(2\) and height \(3\), so its area is \(\frac{1}{2} \cdot 2 \cdot 3 = 3\) square units.

a) \(f(x) = -1.5x + 3\) b) \(3\) square units

- Use the two intercepts to calculate the slope of \(f\). - Keep the y-intercept fixed while changing the slope as described. - After finding both x-intercepts, compare their x-coordinates.

1. The slope of \(f\) is \(m = \frac{0 - (-2)}{3 - 0} = \frac{2}{3}\). With y-intercept \(-2\), \(f: y = \frac{2}{3}x - 2\). 2. Doubling the slope gives \(m_p = \frac{4}{3}\). Keeping the y-intercept \(-2\) gives \(p: y = \frac{4}{3}x - 2\). 3. For the x-intercept of \(p\), solve \(0 = \frac{4}{3}x - 2\). This gives \(x = 1.5\), so the x-intercept is \((1.5, 0)\). 4. The x-coordinate of the intercept changed from \(3\) to \(1.5\), so it was cut in half when the slope was doubled while the y-intercept stayed fixed.

a) \(m = \frac{2}{3}\), and \(f: y = \frac{2}{3}x - 2\) b) \(p: y = \frac{4}{3}x - 2\) c) \((1.5, 0)\). Its x-coordinate is half the x-coordinate of the x-intercept of \(f\).

- Isolate \(y\) to rewrite the equation in slope-intercept form. - Which coordinate equals zero at each axis intercept? - Because \(P(k, 2k)\) lies on the line, its coordinates must satisfy the equation.

1. Solve for \(y\): \(0.2y = -0.4x + 2\), so \(y = -2x + 10\). 2. For the y-intercept, set \(x = 0\), giving \((0, 10)\). For the x-intercept, set \(y = 0\): \(0 = -2x + 10\), so \(x = 5\), giving \((5, 0)\). 3. Substitute \(x = k\) and \(y = 2k\) into the original equation: \(0.4k + 0.2(2k) = 2\). Then \(0.8k = 2\), so \(k = 2.5\).

a) \(y = -2x + 10\) b) x-intercept: \((5, 0)\); y-intercept: \((0, 10)\) c) \(k = 2.5\)

- What does it mean mathematically for a point to satisfy an equation? - How can you rearrange an equation step by step to isolate one variable? - What sign rule is especially important when substituting negative numbers? - Recall the general form of a linear equation.

1. For \(P_1\), \(12 \cdot 0.5 - 4 \cdot 1 = 2 \neq 6\), so it is not a solution. For \(P_2\), \(12 \cdot 1 - 4 \cdot 1.5 = 6\), so it is a solution. For \(P_3\), \(12 \cdot 0 - 4 \cdot (-1.5) = 6\), so it is a solution. 2. For \(Q\), \(24 - 4y = 6\), so \(y = 4.5\). 3. For \(R\), \(12x - 4 \cdot (-3) = 6\), so \(12x = -6\) and \(x = -0.5\). 4. Solving for \(y\) gives \(-4y = -12x + 6\), so \(y = 3x - 1.5\).

a) \(P_2\) and \(P_3\) satisfy the equation; \(P_1\) does not. b) \(y = 4.5\) c) \(x = -0.5\) d) \(y = 3x - 1.5\)

- Substitute the coordinates of \(S\) to find the unknown coefficient. - What coordinate is zero at each axis intercept? - Rewrite both equations in slope-intercept form to compare their slopes and intercepts.

1. Substitute \((2, 3)\): \(2a + 4 \cdot 3 = 20\). Thus, \(2a + 12 = 20\), so \(a = 4\). 2. The equation is \(4x + 4y = 20\). Setting \(y = 0\) gives \(x = 5\), so the x-intercept is \((5, 0)\). Setting \(x = 0\) gives \(y = 5\), so the y-intercept is \((0, 5)\). 3. Replacing \(20\) with \(10\) gives \(4x + 4y = 10\), or \(y = -x + 2.5\). The slope remains \(-1\), while the y-intercept decreases from \(5\) to \(2.5\), so the line shifts parallel to itself.

a) \(a = 4\) b) x-intercept: \((5, 0)\); y-intercept: \((0, 5)\) c) The line remains parallel to the original line and shifts so that its y-intercept is \(2.5\) instead of \(5\).

- How can you isolate \(y\) step by step on one side of the equation? - Where is the slope located in the form \(y = mx + b\)? - What must you do to check whether a point satisfies an equation? - Remember the rules for expanding expressions with parentheses.

1. Expand and solve for \(y\): \(4y - 4 + 2x = 8\), so \(4y = -2x + 12\), and \(y = -0.5x + 3\). 2. Therefore, \(m = -0.5\) and \(b = 3\). 3. At \(x = 6\), the line has \(y = -0.5 \cdot 6 + 3 = 0\). Since \(-1 \neq 0\), \(P(6, -1)\) is not on the line.

a) \(y = -0.5x + 3\) b) \(m = -0.5\), \(b = 3\) c) No. \(P(6, -1)\) is not on the line.

- If \(x = -4\) is a zero, what must \(f(-4)\) equal? - What kind of graph results when the slope is \(0\)? - Can a horizontal line above the x-axis cross it?

1. For part a, use \(f(-4) = 0\): \(-4a + 6 = 0\). Then \(-4a = -6\), so \(a = 1.5\). 2. For part b, let \(a = 0\). Then \(f(x) = 6\), a horizontal line above the x-axis. 3. Since \(f(x)\) is never \(0\) when \(a = 0\), the function has no zero in that case.

a) \(a = 1.5\) b) Yes. For \(a = 0\), \(f(x) = 6\), so the function has no zero.

- Find the slope from the two given points. - Use either point to find the y-intercept. - What function value defines an x-intercept? - What must \(f(0)\) equal for the graph to pass through the origin?

1. The slope is \(m = \frac{2 - 5}{-2 - 4} = 0.5\). 2. Substitute \((4, 5)\) into \(y = 0.5x + b\): \(5 = 0.5 \cdot 4 + b\), so \(b = 3\). Thus \(f(x) = 0.5x + 3\). 3. For the x-intercept, solve \(0 = 0.5x + 3\). This gives \(x = -6\), so the intercept is \((-6, 0)\). 4. A line passes through the origin only if \(f(0) = 0\). Here \(f(0) = 3\), so the graph does not pass through the origin.

a) \(f(x) = 0.5x + 3\) b) \((-6, 0)\) c) No. Since \(f(0) = 3\), the graph does not pass through the origin.

- Pay close attention to signs when finding the slope. - What is the y-coordinate at an x-intercept? - To test a point, substitute its x-coordinate and compare the result with its y-coordinate.

1. Find the slope: \(m = \frac{-5 - 7}{4 - (-2)} = -2\). 2. Substitute \((-2, 7)\) into \(y = -2x + b\): \(7 = -2 \cdot (-2) + b\), so \(b = 3\). Thus \(f(x) = -2x + 3\). 3. For the x-intercept, solve \(0 = -2x + 3\). This gives \(x = 1.5\), so the intercept is \((1.5, 0)\). 4. Check \(R\): \(f(1) = -2 \cdot 1 + 3 = 1\). This matches the y-coordinate of \(R\), so \(R\) lies on the line.

a) \(f(x) = -2x + 3\) b) \((1.5, 0)\) c) Yes. Since \(f(1) = 1\), \(R(1, 1)\) lies on the line.

- What y-value defines an x-intercept? - What do parallel lines have in common about their slopes? - After finding the equation of \(h\), evaluate it at \(x = 0\).

1. For the x-intercept of \(g\), solve \(0 = -2x + 4\). This gives \(x = 2\), so the intercept is \((2, 0)\). 2. Since \(h\) is parallel to \(g\), its slope is \(-2\). 3. Substitute \((1, 1)\) into \(y = -2x + b\): \(1 = -2 \cdot 1 + b\), so \(b = 3\). Thus \(h(x) = -2x + 3\). 4. Since \(h(0) = 3\), the line does not pass through the origin.

a) \((2, 0)\) b) \(h(x) = -2x + 3\) c) No. \(h(0) = 3\), not \(0\).

- What equation do you solve to find an x-intercept? - Which of two x-values lies farther to the right on the x-axis? - Where can you read the y-intercept directly from \(y = mx + b\)?

1. For \(g\), the y-intercept is \((0, 6)\). To find the x-intercept, solve \(-1.5x + 6 = 0\), which gives \(x = 4\). Thus the x-intercept is \((4, 0)\). 2. For \(h\), the y-intercept is \((0, -3)\). To find the x-intercept, solve \(0.5x - 3 = 0\), which gives \(x = 6\). Thus the x-intercept is \((6, 0)\). 3. Since \(6 > 4\), the graph of \(h\) crosses the x-axis farther to the right. 4. The y-intercept of \(g\) is \(6\), which is positive, while the y-intercept of \(h\) is \(-3\).

a) \(g\): x-intercept \((4, 0)\), y-intercept \((0, 6)\); \(h\): x-intercept \((6, 0)\), y-intercept \((0, -3)\) b) \(h\), because \(6 > 4\). c) \(g\)

- What does it mean for two functions to have the same zero? - First find the x-value that makes \(p(x)\) equal to \(0\). - Use that x-value in \(k(x)\) to solve for \(a\).

1. Find the zero of \(p\): \(2x - 8 = 0\), so \(x = 4\). 2. Since \(k\) has the same zero, \(k(4) = 0\). 3. Substitute \(x = 4\): \(4a + 10 = 0\). 4. Then \(4a = -10\), so \(a = -2.5\).

\(a = -2.5\)

- Find the slope from the two given points. - Use slope-intercept form to find the y-intercept. - Which coordinate is zero at each axis intercept?

1. The slope is \(m = \frac{4 - 1}{2 - (-4)} = 0.5\). 2. Substitute \((2, 4)\) into \(y = 0.5x + b\): \(4 = 0.5 \cdot 2 + b\), so \(b = 3\). Thus the line is \(y = 0.5x + 3\). 3. The y-intercept is \((0, 3)\). 4. For the x-intercept, solve \(0 = 0.5x + 3\). This gives \(x = -6\), so the x-intercept is \((-6, 0)\).

The x-intercept is \((-6, 0)\), and the y-intercept is \((0, 3)\).

- Find the slope of \(g\) from its two intercepts. - Change the slope as described for \(h\). - Use the shared x-intercept to find the new y-intercept.

1. The slope of \(g\) is \(m_g = \frac{0 - (-2)}{4 - 0} = 0.5\). 2. Twice that slope is \(m_h = 1\). 3. Since \(h\) passes through \((4, 0)\), substitute into \(y = x + b_h\): \(0 = 4 + b_h\), so \(b_h = -4\). 4. Therefore, \(h: y = x - 4\), and its y-intercept is \((0, -4)\).

\(h: y = x - 4\), with y-intercept \((0, -4)\).

- What is the general slope-intercept form of a linear equation? - How can a known slope and one point be used to find the missing parameter? - What is the x-coordinate of every point on the y-axis? - What is the y-coordinate of every point on the x-axis?

1. Substitute the slope and point into \(y = mx + b\): \(1 = -0.75 \cdot 4 + b\), so \(b = 4\). Thus \(g\) has equation \(y = -0.75x + 4\). 2. At the y-intercept, \(x = 0\), so the point is \((0, 4)\). 3. At the x-intercept, \(y = 0\): \(0 = -\frac{3}{4}x + 4\). Solving gives \(x = \frac{16}{3}\), so the point is \((\frac{16}{3}, 0)\).

The y-intercept is \((0, 4)\), and the x-intercept is \((\frac{16}{3}, 0)\).

- How do you find the vertical change divided by the horizontal change between two points? - Once you know the slope, use either point to find the y-intercept. - What does the size of a positive slope tell you about a line's steepness?

1. The slope is \(m = \frac{10 - 4}{3 - 1} = 3\). 2. Substitute \((1, 4)\) into \(y = 3x + b\): \(4 = 3 \cdot 1 + b\), so \(b = 1\). Thus \(f(x) = 3x + 1\). 3. The original slope is \(3\). Halving it gives \(1.5\). With the same y-intercept, the new line still increases but is less steep.

a) \(f(x) = 3x + 1\) b) \(m = 3\) c) The new slope is \(1.5\), so the line still rises from left to right but is less steep.

- How can you find the slope from two points in the table? - What is the function value at an x-intercept? - What does the parameter \(b\) control in the graph of \(y = mx + b\)?

1. The slope is \(m = \frac{1 - (-5)}{1 - (-2)} = \frac{6}{3} = 2\). 2. Substitute \((1, 1)\) into \(y = 2x + b\): \(1 = 2 \cdot 1 + b\), so \(b = -1\). Thus \(f(x) = 2x - 1\). 3. For the x-intercept, solve \(2x - 1 = 0\). This gives \(x = 0.5\), so the x-intercept is \((0.5, 0)\). 4. Increasing \(b\) by \(3\) changes the equation to \(y = 2x + 2\). The slope stays the same, so the graph shifts vertically upward by \(3\) units.

a) \(f(x) = 2x - 1\) b) \((0.5, 0)\) c) The graph shifts upward by \(3\) units and remains parallel to the original line.

- For a reflection across the x-axis, compare each output of one function with the opposite of the other function's output. - Which points stay fixed when a graph is reflected across the y-axis? - Think about how replacing \(x\) with \(-x\) affects the slope of a linear function.

1. Two function graphs are reflections across the x-axis when \(g(x) = -f(x)\) for every \(x\). 2. Here, \(-f(x) = -\left(\frac{2}{3}x + 2\right) = -\frac{2}{3}x - 2 = g(x)\). Therefore, the graphs are reflections across the x-axis. 3. A reflection across the y-axis replaces \(x\) with \(-x\), so \(h(x) = f(-x)\). 4. Then \(h(x) = \frac{2}{3}(-x) + 2 = -\frac{2}{3}x + 2\). The y-intercept stays \(2\), while the slope changes from \(\frac{2}{3}\) to \(-\frac{2}{3}\).

a) Yes. Since \(g(x) = -f(x)\) for every \(x\), the graphs are reflections across the x-axis. b) \(h(x) = -\frac{2}{3}x + 2\). The y-intercept remains \(2\), and the slope changes sign.

- Substitute each value of \(k\) before simplifying. - After simplifying, identify whether x or y is held constant. - Use the orientation of the resulting line to reason about function status and slope.

1. For \(k=1\), the equation becomes \(0x+3y=6\), so \(y=2\). This is a horizontal line parallel to the x-axis. 2. For \(k=-2\), the equation becomes \(-3x+0y=6\), so \(x=-2\). This is a vertical line. It is not the graph of \(y=f(x)\) because the input \(x=-2\) corresponds to infinitely many y-values. 3. The line \(y=2\) has slope \(0\). The line \(x=-2\) has undefined slope.

1) \(y=2\); horizontal and parallel to the x-axis 2) \(x=-2\); vertical and not a function of x 3) For \(k=1\), \(m=0\); for \(k=-2\), the slope is undefined.

- Use the slope formula with the two points. - Substitute one point into \(y=mx+b\) to find \(b\). - The sign and magnitude of \(m\) describe the line's direction and rate of change. - Changing only \(b\) creates a vertical translation.

1. The slope is \(m=\frac{-7-5}{4-(-2)}=\frac{-12}{6}=-2\). 2. Substitute \(A(-2, 5)\) into \(y=-2x+b\): \(5=-2\cdot(-2)+b\), so \(b=1\). 3. Since \(m=-2\), the line falls \(2\) units for every \(1\)-unit increase in \(x\). Its graph decreases from left to right. 4. Decreasing \(b\) by \(3\) changes the equation to \(y=-2x-2\), translating the graph downward \(3\) units without changing its slope.

1. \(m=-2\), \(b=1\) 2. The line decreases, falling \(2\) units for each \(1\)-unit increase in \(x\). 3. The graph shifts downward \(3\) units.

- Where does each displayed line cross the y-axis? - Use two grid points on a line to find its slope. - What must be true about the slopes of parallel lines? - For a line through two points, find the slope first and then use one point to find the y-intercept.

1. Line \(f\) crosses the y-axis at \((0, 2)\). Using \((0, 2)\) and \((3, 3)\), its slope is \(\frac{1}{3}\). Thus \(f(x) = \frac{1}{3}x + 2\). 2. Line \(g\) crosses the y-axis at \((0, -1)\). Using \((0, -1)\) and \((2, -2)\), its slope is \(-\frac{1}{2}\). Thus \(g(x) = -\frac{1}{2}x - 1\). 3. Because \(h\) is parallel to \(f\), its slope is \(\frac{1}{3}\). Using \(P(-3, 3)\), \(3 = \frac{1}{3} \cdot (-3) + b\), so \(b = 4\). Therefore, \(h(x) = \frac{1}{3}x + 4\). 4. The slope through \(P(-3, 3)\) and \(Q(1, -1)\) is \(\frac{-1 - 3}{1 - (-3)} = -1\). Using \(Q\), \(-1 = -1 + b\), so \(b = 0\). Therefore, \(k(x) = -x\).

a) \(f(x) = \frac{1}{3}x + 2\) and \(g(x) = -\frac{1}{2}x - 1\) b) \(h(x) = \frac{1}{3}x + 4\) c) \(k(x) = -x\)

- Choose two grid points on each line. - Use rise over run to find the slope. - Read the y-intercept where the line crosses the vertical axis. - Clear fractions before rearranging into \(Ax + By = C\).

1. Line \(a\) passes through \((0, -2)\) and \((3, 0)\). Its slope is \(m = \frac{0 - (-2)}{3 - 0} = \frac{2}{3}\), and its y-intercept is \(-2\). Thus \(y = \frac{2}{3}x - 2\). Multiplying by \(3\) and rearranging gives \(2x - 3y = 6\). 2. Line \(b\) passes through \((0, 3)\) and \((3, 0)\). Its slope is \(-1\), and its y-intercept is \(3\). Thus \(y = -x + 3\), which is equivalent to \(x + y = 3\).

Line \(a\): \(y = \frac{2}{3}x - 2\) and \(2x - 3y = 6\); line \(b\): \(y = -x + 3\) and \(x + y = 3\)

- Recover each line equation from two vertices on the same segment. - Keep the given interval convention aligned with the segment formulas. - For \(f(x+2)\), transform both the formulas and the interval boundaries.

1. The first segment goes through \((-3,-1)\) and \((-1,2)\), so its slope is \(\frac{3}{2}\) and its rule is \(\frac{3}{2}x+\frac{7}{2}\). 2. The second segment goes through \((-1,2)\) and \((2,0)\), so its slope is \(-\frac{2}{3}\) and its rule is \(-\frac{2}{3}x+\frac{4}{3}\). 3. The third segment goes through \((2,0)\) and \((4,1)\), so its slope is \(\frac{1}{2}\) and its rule is \(\frac{1}{2}x-1\). 4. Therefore, \(f\) uses those three rules on \([-3,-1)\), \([-1,2)\), and \([2,4]\), respectively. 5. For \(h(x)=f(x+2)\), each breakpoint shifts left \(2\) units, so the intervals become \([-5,-3)\), \([-3,0)\), and \([0,2]\). 6. Substituting \(x+2\) into the three branch formulas gives \(\frac{3}{2}x+\frac{13}{2}\), \(-\frac{2}{3}x\), and \(\frac{1}{2}x\). The new domain is \([-5,2]\).

a) \(f(x)=\begin{cases}\frac{3}{2}x+\frac{7}{2}&-3\le x<-1\\-\frac{2}{3}x+\frac{4}{3}&-1\le x<2\\\frac{1}{2}x-1&2\le x\le4\end{cases}\) b) \(h(x)=\begin{cases}\frac{3}{2}x+\frac{13}{2}&-5\le x<-3\\-\frac{2}{3}x&-3\le x<0\\\frac{1}{2}x&0\le x\le2\end{cases}\), domain \([-5,2]\). Every breakpoint moves left \(2\) units.

- Compare the graph's rise-and-run pattern with the coefficient of \(x\). - Check the graph where \(x=0\). - If the slope is already correct, decide what change can alter only the intercept.

1. The shown line has slope \(2\), but it crosses the y-axis at \(1\). 2. Therefore, the shown graph represents \(y=2x+1\), not \(y=2x-1\). 3. The correct graph should be parallel to the shown line but shifted down \(2\) units so that it crosses the y-axis at \(-1\).

The slope is correct, but the y-intercept is wrong. The correct line is parallel to the shown line and crosses the y-axis at \(-1\), which is \(2\) units lower.

- Parallel lines can be identified by comparing their steepness and direction. - For y-intercepts, inspect where each line crosses the vertical axis. - For x-intercepts, inspect where each line crosses the horizontal axis.

1. Lines \(f\) and \(g\) have the same slope \(\frac{1}{2}\), so they are parallel. 2. Lines \(f\) and \(h\) both cross the y-axis at \(2\), so they have the same y-intercept. 3. Line \(g\) reaches the x-axis at \((2,0)\), and line \(h\) also reaches it at \((2,0)\). Thus, \(g\) and \(h\) have the same x-intercept.

a) \(f\) and \(g\), because they have the same slope and are parallel. b) \(f\) and \(h\), because both cross the y-axis at \(2\). c) \(g\) and \(h\), because both cross the x-axis at \(x=2\).

- First locate the value the graph should have at time zero. - Then decide whether the line should rise or fall and how quickly. - The physical model stops when the volume reaches zero.

1. At \(t=0\), the tank contains \(18\,\text{gal}\), so the graph must begin at \((0,18)\). 2. The volume decreases by \(3\,\text{gal}\) per minute, so the graph must have slope \(-3\). 3. Graph b) begins at \(18\) gallons and decreases to \(0\) gallons in \(6\) minutes, matching \(V(t)=18-3t\) on \([0,6]\).

b). It starts at \(18\,\text{gal}\) when \(t=0\) and falls by \(3\,\text{gal}\) for each additional minute.

- Think about how changing only the y-intercept shifts a line vertically. - If solving an equation for \(x\) always gives one real value, what does that tell you about the number of zeros? - Consider how the value of a negative multiple changes when the variable increases.

1. For part a, solve \(2x + c = 0\): \(2x = -c\), so \(x = -\frac{c}{2}\). This gives exactly one real value of \(x\) for every real \(c\), so each function has exactly one zero. 2. For part b, the zero is \(x = -\frac{c}{2}\). As \(c\) increases, \(-\frac{c}{2}\) decreases, so the zero moves to the left on the x-axis. 3. For part c, set \(-4.5 = -\frac{c}{2}\). Multiplying both sides by \(-2\) gives \(c = 9\).

a) \(x = -\frac{c}{2}\); therefore every function in the family has exactly one zero. b) As \(c\) increases, the zero moves left because \(-\frac{c}{2}\) decreases. c) \(c = 9\)

- A nonhorizontal line through \(P\) can be written using a chosen nonzero slope. - To avoid Quadrant I, examine what your line does for positive \(x\)-values. - For part b), use the fact that a linear-function graph extends indefinitely in both directions.

1. One valid choice is \(f(x)=-x-6\). It passes through \(P\) because \(-(-2)-6=-4\). For every \(x>0\), \(f(x)=-x-6<0\), so the graph cannot enter Quadrant I. 2. No line can lie entirely in Quadrant III. A linear-function graph extends indefinitely for all real \(x\). A horizontal line below the x-axis enters Quadrant IV, and every nonhorizontal line eventually extends into another quadrant as \(x\) moves without bound.

a) One possible answer is \(f(x)=-x-6\). For every \(x>0\), \(f(x)<0\), so the graph never enters Quadrant I. b) No. A line extends indefinitely in both horizontal directions and cannot remain entirely inside Quadrant III.

- What do graphs of constant functions look like? - How far apart must two horizontal lines be for the height to be 3? - Find each vertex by solving where two boundary lines intersect. - What symmetry would make the trapezoid isosceles?

1. Since \(f_3(x) = 0\) is the x-axis and the trapezoid has height 3, the parallel upper base must be \(f_4(x) = 3\). 2. On \(y = 0\), \(0 = 2x + 4\) gives \(x = -2\), and \(0 = -2x + 4\) gives \(x = 2\). The lower vertices are \((-2, 0)\) and \((2, 0)\). 3. On \(y = 3\), \(3 = 2x + 4\) gives \(x = -0.5\), and \(3 = -2x + 4\) gives \(x = 0.5\). The upper vertices are \((-0.5, 3)\) and \((0.5, 3)\). 4. The two slanted sides have slopes \(2\) and \(-2\), so the figure is symmetric about the y-axis and is an isosceles trapezoid.

\(f_4(x) = 3\). The vertices are \((-2, 0)\), \((2, 0)\), \((0.5, 3)\), and \((-0.5, 3)\).

- Compare several choices of \(n\), including \(0\), while holding the slope fixed. - Consider separately what a nonzero slope and a zero slope imply about x-axis intersections. - Recall what line is determined by two distinct points that both lie on the x-axis.

1. For part a, with \(m = 1.5\), a zero satisfies \(1.5x + n = 0\). Thus \(x = -\frac{n}{1.5}\), which gives exactly one zero for every real value of \(n\). The claim is true in this case. 2. For part b, when \(m = 0\), the equation is \(y = n\). If \(n \neq 0\), the horizontal line does not meet the x-axis, so there are no zeros. If \(n = 0\), the line is the x-axis itself, so every real \(x\) is a zero. The number of zeros can therefore change from none to infinitely many. 3. For part c, two distinct points determine exactly one line. If a line intersects the x-axis at two different points, it must be the x-axis. Then it has infinitely many intersection points with the x-axis, not exactly two.

a) No. When \(m = 1.5\), there is exactly one zero for every real \(n\). b) The claim is false when \(m = 0\): if \(n \neq 0\), there are no zeros; if \(n = 0\), there are infinitely many zeros. c) No. A line through two different points on the x-axis must be the x-axis itself, so it has infinitely many such intersection points.

- What mathematical operation is described by multiplying every y-coordinate by \(3\)? - In an equation of the form \(y = mx\), where can you see the slope? - For the general case, express the new y-coordinate in terms of the old one. - Check whether the transformed equation still has the form of a line through the origin.

1. Multiplying every output by \(3\) gives \(h(x) = 3f(x)\). Therefore, \(h(x) = 3(1.5x) = 4.5x\). 2. The slope changes from \(1.5\) to \(4.5\). The new slope is three times the original slope, so the graph is steeper. 3. Start with \(y = mx\). Multiplying each output by \(k\) gives \(y_{new} = k(mx) = (km)x\). This is still of the form \(y = m'x\), so it is a line through the origin with slope \(m' = km\).

a) \(h(x) = 4.5x\) b) The slope changes from \(1.5\) to \(4.5\), so it is multiplied by \(3\) and the graph becomes steeper. c) The new equation is \(y = (km)x\), so the new slope is \(km\).