Math assignment Essay

Custom Student Mr. Teacher ENG 1001-04 12 July 2017

Math assignment

A linear function is formed by the slope-intercept form y = mx + b, where m is the gradient/slope, and b is the y-intercept. The greater gradient, the steeper the line will be. The gradient can be calculated through taking ? y/? x. Then you will have to use two points on each line. The y-intercept of my first linear function graph, the red line, is 2, due to that b = 2. The gradient is + 3, which also can be written 3/1. To find out that, it is possible to use the rise/run method. That means that you already know one point on the graph, the y-intercept + 2.

From that point you can go upwards (rise) three units, and one unit to the right (run). You could also use the ? y/? x method to calculate the gradient. Then you take two co-ordinates on the graph. In my example I have chosen to take (-1, -1) and (0,2). I then take (2-(-1)) / (0-(-1)) = 3/1 which gives me the same answer (y2-y1 / x2-x1). In my second linear function graph, the gradient is negative, which means that my line will be moving downwards from left to right. The y-intercept is – 4 and the gradient is – 2.

To calculate the slope I use the co-ordinates (-2,0) and (-1, -2). I use the ? y/? x method once again. (-2-0) / (-1-(-2)) = -(2/1). Absolute Value Function: In an absolute value function, the graph is determined from the sign that is before the absolute value signs, since what is inside always will be positive after removing the absolute value signs. The characteristics of an absolute value graph are that it is a V-shape, either positive that opens upwards, or negative that opens downwards, as shown in my examples.

The function is often written Y = | x – b|. The vertex is (b, 0). Since the b is inside the absolute value symbols, it is always positive, but there can be a negative sign before those absolute symbols, that change it, as in my second function. The y-intercept for an absolute value function is b, and it is not to be forgotten that b is inside the absolute value symbols. So, you calculate the graph through using the three points that you have got, the y-intercept, the vertex, and the point, symmetric to the y-intercept. Quadratic Functions:

Quadratic functions are written in the standard form y = ax2 + bx + c. This type of graph is like a parabola that opens upward, or downward. If a < 0, the parabola opens downward, and if a > 0, the parabola opens upward. The graph will be more narrow the greater the value of a is. So, as you can see in my graph, the value of a is greater in my first graph, than in my second graph, and the parabola is therefore more narrow.

The axis of symmetry can be calculated through taking x = (-b) / (2a), so in my case for the first graph, it would be x = (-2) / (2*3) which equals which you can see is correct. The axis of symmetry is used to see that the graph is correctly drawn. You can fold the paper there, and the graph should look exactly the same on both sides, if you have done it right. The vertex, which has the maximum value of the graph if it opens downward, and minimium value of the graph if it opens upward, ( -b / (2a), y). In the first of my graphs, it would be ( (-2) / (2*3), -(2 ? ) ). The y-intercept in a quadratic function is always the value of c.

The easiest way to draw a quadratic function is through using a Graphing Calculater, or a table of values where you enter the domain and range. Quadratic functions can, but do not have to have a solution. The solutions are the x-intercepts, and can easily be calculated with the formula x = (-b i?? V (b2 -4ac)) / 2a, and that will give us the solutions. You have to use this formula twice, once with the positive sign before the square root of, and once with the negative sign.

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  • University/College: University of Arkansas System

  • Type of paper: Thesis/Dissertation Chapter

  • Date: 12 July 2017

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