Determining Slope and y-Intercept Videos - Free Educational Videos for Students in K - 12

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Inverse Functions | MathHelp.com


By MathHelp.com

In this example, we’re given a relation in the form of a chart, and we’re asked to find the inverse of the relation, then graph the relation and its inverse. To find the inverse of a relation, we simply switch the x and y values in each point. In other words, the point (1, -4) becomes (-4, 1), the point (2, 0) becomes (0, 2), the point (3, 1) becomes (1, 3), and the point (6, -1) becomes (-1, 6). Next, we’re asked to graph the relation and its inverse, so let’s first graph the relation. Notice that the relation contains the points (1, -4,), (2, 0), (3, 1), and (6, -1). And the inverse of the relation contains the points (-4, 1), (0, 2), (1, 3), and (-1, 6). Finally, it’s important to understand the following relationship between the graph of a relation and its inverse. If we draw a diagonal line through the coordinate system, which is the line that has the equation y = x, notice that the relation and its inverse are mirror images of each other in this line. In other words, the inverse of a relation is the reflection of the original relation in the line y = x.

Systems of Three Equations | MathHelp.com


By MathHelp.com

Here we’re asked to graph the following function and use the horizontal line test to determine if it has an inverse. And if so, find the inverse function and graph it. So let’s start by graphing the given function, f(x) = 2x – 4, and remember that f(x) is the same as y, so we can rewrite the function as y = 2x – 4. Now, we simply graph the line y = 2x – 4, which has a y-intercept of -4, and a slope of 2, or 2/1, so we go up 2 and over 1, plot a second point and graph our line, which we’ll call f(x). Next, we’re asked to use the horizontal line test to determine if the function has an inverse. Since there’s no way to draw a horizontal line that intersects more than one point on the function, the function does have an inverse. So we need to find the inverse and graph it. To find the inverse, we switch the x and the y in original function, y = 2x – 4, to get x = 2y – 4. Next, we solve for y, so we add 4 to both sides to get x + 4 = 2y, and divide both sides by 2 to get 1/2x + 2 = y. Next, let’s flip our equation so that y is on the left side, and we have y = 1/2x + 2. Finally, we replace y with the notation that we use for the inverse function of f, as shown here. And remember that we’re asked to graph the inverse as well, so we graph y = ½ x + 2. Our y-intercept is positive 2, and our slope is ½, so we go up one and over 2, plot a second point, graph the line, and label it as the inverse function of f. Notice that the graph of the inverse function is a reflection of the original function in the line y = x.

Difference of Two Cubes | MathHelp.com


By MathHelp.com

To solve the given system of inequalities, we start by graphing the associated equation for each inequality. In other words, we graph y equals -1/5 x +1 and y equals 3x + 2. So, for the first inequality, we start with our y-intercept of positive 1, up 1 unit on the y-axis. From there, we take our slope of -1/5, so we go down 1 and to the right 5, and plot a second point. Now, notice that our inequality uses a “less than” sign. This means that we draw a dotted line connecting the points, rather than a solid line. It’s important to understand that if we have a greater than sign or a less than sign, we use a dotted line, and if we have a greater than or equal to sign or a less than or equal to sign, we use a solid line. Pay close attention to this idea when drawing your lines. Students often carelessly use a solid line when they should use a dotted one, and vice-versa. Next, let’s take a look at our second inequality, which has a y-intercept of positive 2, up 2 units on the y-axis. From there, we take our slope of 3, or 3 over 1, so we go up 3 and to the right 1, and plot a second point. And notice that this inequality uses a “greater than or equal to” sign, so we connect the points with a solid line, rather than a dotted line. Next, we need to determine which side of each of these lines to shade on the graph. To determine which side of our first line to shade, we use a test point on either side of the first line. The easiest test point to use is (0, 0), so we plug a zero into our first inequality for both x and y, and we have 0 is less than -1/5 times 0 + 1, which simplifies to 0 is less than 0 + 1, or 0 is less than 1. Notice that 0 is less than 1 is a true statement. This means that our test point, (0, 0), is a solution to the first inequality, so we shade in the direction of (0, 0) along our first boundary line. Next, we determine which side of our second line to shade by using a test point on either side of the second line, such as (0, 0). Plugging a zero into our second inequality for both x and y, we have 0 equal to 3 times 0 + 2, which simplifies to 0 equal to 0 + 2, or 0 equal to 2. Notice that 0 equal to 2 is a false statement. This means that our test point, (0, 0), is a not solution to the inequality, so we shade away from (0, 0) along our second boundary line. Finally, it’s important to understand that the solution to this system of inequalities is represented by the part of the graph where the two shaded regions overlap, which in this case is in the lower left. In other words, any point that lies in this part of the graph is a solution to the given system of inequalities. Note that the points along the dotted boundary line of this region are not solutions to the system, but the points along the solid boundary line of this region are solutions to the system.

ALL OF GRADE 9 MATH IN 60 MINUTES!!! (exam review part 2)


By Lumos Learning

Here is a great exam review video reviewing all of the main concepts you would have learned in the MPM1D grade 9 academic math course. The video is divided in to 3 parts. This is part 2: Linear Relations. In this video you will review everything there is to know about y=mx+b, scatterplots, and distance time graphs.

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05 - Quadratic Systems of Equations (With Lines, Circles, Ellipses, Parabolas & Hyperbolas)


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