Pay stubs & time sheets
NOTES: A quick overview of advanced graphing In order to do the procedures here, you’ll need to be familiar with using negative numbers, applying the distributive property, and solving linear equations. For the end you’ll also need to be able to calculate slope. NOTE: If a question asks you to write “the equation of a line,” it usually means an equation in y = mx + b form (slope-intercept form).
How do I write the equation of a line from a point & the slope? There are 2 common methods. Pick whichever makes more sense to you. EXAMPLE: Write the equation of a line that has a slope of –4 and that passes through the point (2, 5). METHOD METHOD1: 1: point-slope point-slopemethod method This method is conceptually pretty straightforward. You just plug the coordinates for the given point and the given slope into the point-slope form of the equation. Then you manipulate the equation until it has the y = mx + b form. The downside is that you need to remember the point-slope formula or look it up. Also, the algebra is a little trickier than in method 2.
1. Write down the point-slope formula: y y1 = m(x x1) (This is on your formula sheet. Notice that it’s basically just the slope formula rearranged.) m is the slope. You already have that: –4 (x1, y1) x1 and y1 are the coordinates of the given point: (2, 5) 2. Plug in the known values for m, x1 and y1. (Make sure not to touch the x and y in the formula—just the x1 and y1.)
y y1 = m(x x1) y 5 = 4(x 2) continued on next page…
© D. Stark 2017
10/2/2017
OVERVIEW: advanced graphing
1
3. Use the distributive property and your knowledge of how to solve equations to get the equation into y = mx + b form: y 5 = 4(x 2) y 5 = 4x + 8 +5 +5 y = 4x + 13 METHOD2: 2: find findbbmethod method(slope-intercept (slope-interceptmethod) method) METHOD This method is probably a bit harder to remember. But the algebra is simpler, and you get to stick with the more familiar y = mx + b. The idea here is that you already know the slope (m) but not the y-intercept (b). You use the one point you do know to find b.
1. Write down the slope-intercept formula: y = mx + b m is the slope. You already have that: –4 You already know one solution: (2, 5). That’s when x = 2 and y = 5. What you don’t know is b. That’s what you need to find. 2. Plug the known values for m, x, and y. [Note that in this method you DO plug in for the x and y in the formula.] y = mx + b 5 = –4(2) + b 3. Use your knowledge of solving equations to solve for b. 5 = –4(2) + b 5 = –8 + b +8 +8 13 = b or b = 13 4. Now you know the slope (–4) and the y-intercept (13) so you can write the equation in y = mx + b form: y = –4x + 13
How do I write the equation of a line from 2 points? 1. First, find the slope, using the slope formula: 𝑚 =
𝑦2 −𝑦1 𝑥2 −𝑥1
2. Then you’re back to the case of having a point and the slope. Use one of the methods above with either of the 2 given points—it doesn’t matter which.
© D. Stark 2017
10/2/2017
OVERVIEW: advanced graphing
2
How do I graph a function with a table of values? 1. Pick at least 3 convenient values for x, and evaluate the function for those values, putting the results in a table. Sometimes you’ll see a horizontal table instead of a vertical one. If the coefficient of x is a fraction (as in f(x) =
2 3
x – 4), it’s best to pick
values for x that are multiples of the denominator so you can don’t have fractions to graph.) Otherwise, you might use –1, 0, 1 or 0, 1, 2. 2. Each row of a vertical table (or each column of a horizontal one) expresses an ordered pair. Plot those ordered pairs on the coordinate plane. The horizontal axis has the independent variable—the one you plug in, which is usually x. The vertical axis has the dependent variable—the one whose value you calculate. That’s usually y but could be labeled f(x), g(x), etc. 3. Draw the appropriate straight or curved line to connect the dots. For the functions we’ll be graphing, you’ll also need to draw arrows at the ends. EXAMPLES: f(x) = 2x – 1 (linear function: straight line) f(x)
f(x) = –
𝟐 𝟑
x+1
(linear function: straight line) x
f(x)
–1 –3
–3
3
0 –1
0
1
1
3 –1
x
f(x)
1
© D. Stark 2017
10/2/2017
f(x)
OVERVIEW: advanced graphing
3
g(x) = x2 – 1 (quadratic function: parabola) g(x)
g(x) = –2x2 + 1 (quadratic function: parabola) x
g(x)
0
–1
–1
0 –1
0
1
1
1
–1
x
g(x)
–1
0
g(x)
Quadratic functions have an x2 term. The graph of a quadratic function is a parabola, like a “U” you can imagine folding in half across the axis of symmetry. The maximum or minimum point of a parabola is the vertex. If the coefficient of x2 is positive, it opens up. If it’s negative, it opens down (just as the ± slope m of a line in y = mx + b form shows it as uphill or downhill). If the coefficient of x2 is large, the parabola is skinny; if it’s a fraction, it’s fat (just as the m value makes a straight line more or less steep). The constant term gives the y-intercept (just as it does in lines: y = mx + b). Quadratics can also be shifted left or right, but we won’t deal with that here. h(x) = x3 – 1 x
h(x)
(cubic function) h(x)
h(x) = –2x3 + 1 x
h(x)
–2 –9
–2 17
–1 –2
–1
3
0 –1
0
1
1
0
1 –1
2
7
2 –15
(cubic function) h(x)
GED® is a registered trademark of the American Council on Education (ACE) and administered exclusively by GED Testing Service LLC under license. This material is not endorsed or approved by ACE or GED Testing Service.
© D. Stark 2017
10/2/2017
OVERVIEW: advanced graphing
4
How do I write the equation of a line from a point & the slope? There are 2 common methods. Pick whichever makes more sense to you. EXAMPLE: Write the equation of a line that has a slope of –4 and that passes through the point (2, 5). METHOD METHOD1: 1: point-slope point-slopemethod method This method is conceptually pretty straightforward. You just plug the coordinates for the given point and the given slope into the point-slope form of the equation. Then you manipulate the equation until it has the y = mx + b form. The downside is that you need to remember the point-slope formula or look it up. Also, the algebra is a little trickier than in method 2.
1. Write down the point-slope formula: y y1 = m(x x1) (This is on your formula sheet. Notice that it’s basically just the slope formula rearranged.) m is the slope. You already have that: –4 (x1, y1) x1 and y1 are the coordinates of the given point: (2, 5) 2. Plug in the known values for m, x1 and y1. (Make sure not to touch the x and y in the formula—just the x1 and y1.)
y y1 = m(x x1) y 5 = 4(x 2) continued on next page…
© D. Stark 2017
10/2/2017
OVERVIEW: advanced graphing
1
3. Use the distributive property and your knowledge of how to solve equations to get the equation into y = mx + b form: y 5 = 4(x 2) y 5 = 4x + 8 +5 +5 y = 4x + 13 METHOD2: 2: find findbbmethod method(slope-intercept (slope-interceptmethod) method) METHOD This method is probably a bit harder to remember. But the algebra is simpler, and you get to stick with the more familiar y = mx + b. The idea here is that you already know the slope (m) but not the y-intercept (b). You use the one point you do know to find b.
1. Write down the slope-intercept formula: y = mx + b m is the slope. You already have that: –4 You already know one solution: (2, 5). That’s when x = 2 and y = 5. What you don’t know is b. That’s what you need to find. 2. Plug the known values for m, x, and y. [Note that in this method you DO plug in for the x and y in the formula.] y = mx + b 5 = –4(2) + b 3. Use your knowledge of solving equations to solve for b. 5 = –4(2) + b 5 = –8 + b +8 +8 13 = b or b = 13 4. Now you know the slope (–4) and the y-intercept (13) so you can write the equation in y = mx + b form: y = –4x + 13
How do I write the equation of a line from 2 points? 1. First, find the slope, using the slope formula: 𝑚 =
𝑦2 −𝑦1 𝑥2 −𝑥1
2. Then you’re back to the case of having a point and the slope. Use one of the methods above with either of the 2 given points—it doesn’t matter which.
© D. Stark 2017
10/2/2017
OVERVIEW: advanced graphing
2
How do I graph a function with a table of values? 1. Pick at least 3 convenient values for x, and evaluate the function for those values, putting the results in a table. Sometimes you’ll see a horizontal table instead of a vertical one. If the coefficient of x is a fraction (as in f(x) =
2 3
x – 4), it’s best to pick
values for x that are multiples of the denominator so you can don’t have fractions to graph.) Otherwise, you might use –1, 0, 1 or 0, 1, 2. 2. Each row of a vertical table (or each column of a horizontal one) expresses an ordered pair. Plot those ordered pairs on the coordinate plane. The horizontal axis has the independent variable—the one you plug in, which is usually x. The vertical axis has the dependent variable—the one whose value you calculate. That’s usually y but could be labeled f(x), g(x), etc. 3. Draw the appropriate straight or curved line to connect the dots. For the functions we’ll be graphing, you’ll also need to draw arrows at the ends. EXAMPLES: f(x) = 2x – 1 (linear function: straight line) f(x)
f(x) = –
𝟐 𝟑
x+1
(linear function: straight line) x
f(x)
–1 –3
–3
3
0 –1
0
1
1
3 –1
x
f(x)
1
© D. Stark 2017
10/2/2017
f(x)
OVERVIEW: advanced graphing
3
g(x) = x2 – 1 (quadratic function: parabola) g(x)
g(x) = –2x2 + 1 (quadratic function: parabola) x
g(x)
0
–1
–1
0 –1
0
1
1
1
–1
x
g(x)
–1
0
g(x)
Quadratic functions have an x2 term. The graph of a quadratic function is a parabola, like a “U” you can imagine folding in half across the axis of symmetry. The maximum or minimum point of a parabola is the vertex. If the coefficient of x2 is positive, it opens up. If it’s negative, it opens down (just as the ± slope m of a line in y = mx + b form shows it as uphill or downhill). If the coefficient of x2 is large, the parabola is skinny; if it’s a fraction, it’s fat (just as the m value makes a straight line more or less steep). The constant term gives the y-intercept (just as it does in lines: y = mx + b). Quadratics can also be shifted left or right, but we won’t deal with that here. h(x) = x3 – 1 x
h(x)
(cubic function) h(x)
h(x) = –2x3 + 1 x
h(x)
–2 –9
–2 17
–1 –2
–1
3
0 –1
0
1
1
0
1 –1
2
7
2 –15
(cubic function) h(x)
GED® is a registered trademark of the American Council on Education (ACE) and administered exclusively by GED Testing Service LLC under license. This material is not endorsed or approved by ACE or GED Testing Service.
© D. Stark 2017
10/2/2017
OVERVIEW: advanced graphing
4
