As discussed in the module, Quadratic equations , this can be solved in three ways:. Completing the square, which is often done to find the vertex and axis of symmetry anyway, is often the most efficient way of laying bare all of the features of the parabola. This method will, of course, work even if the x -intercepts are surds.
Since the right-hand side is always at least 2, the y -values are never zero. Thus this parabola has no x -intercepts. The problem of completing the square for equations of upside-down parabolas is tricky. We can then treat the quadratic in the brackets in the usual way. To find the x -intercepts, we set and so. In each case complete the square and determine the x- and y -intercepts, the axis of symmetr y and the vertex of the parabola. There is a further transformation that results in stretching the arms of the parabola, producing a new parabola that is not congruent to the original one.
Completing the square for non-monic quadratics. The following material should be regarded as extension, since it is tricky and the use of calculus in the senior syllabus can also be used to find the vertex. This is demonstrated in the following example. Find the y -intercept, the axis of symmetr y and the vertex of the parabolas b y completing the square.
Sketch their graphs. The axis of symmetry is a useful line to find since it gives the x -coordinate of the vertex. This gives us the equation of the axis of symmetry and also the x -coordinate and the y -coordinate of the vertex. Find its equation. Symmetry and the x -intercepts.
We have seen that the parabola has an axis of symmetry. In the case when the parabola cuts the x -axis, the x -coordinate of the axis of symmetry lies midway between the two x -intercepts.
Hence the x -coordinate of the vertex is the average of the x -intercepts. We can use this property in some instances to sketch the parabola. Factor if necessary, and sketch, marking intercepts, axis of symmetr y and vertex.
We have emphasized completing the square because it is a such a useful technique and quickly reveals most of the important features of the parabola. If we are given a parabola in factored form, then we can sketch it without expanding and completing the square. Applications involving quadratics. For example, in physics, the displacement of a particle at time with initial velocity and acceleration is given by.
Thus, given the values of u and a , the graph of s against t is a parabola. What is the maximum height that the ball will reach? This is the time at which the ball reaches its maximum height.
The example above is one of a host of problems where we try to find the value of one variable that will minimise or maximise another. In senior mathematics a more powerful technique using differential calculus will be used to achieve this.
A farmer needs to construct a small rectangular paddock using a long wall for one side of the paddock. He has enough posts and wire to erect m of fence. What are the dimensions of the paddock if the fences are to enclose the largest possible area? Let x m be the length of the side perpendicular to the wall. Let A m 2 be the area of the paddock.
Thus the dimensions of the paddock are 50m b y m, and the area is square metres. What is the maximum possible area of the rectangle? What are the coordinates of the vertices of such a rectangle? There is a nice graphical approach to solving this problem.
Thus we can draw the graph as shown. Quadratics are sometimes called equations of degree 2. Equations of general degree are called polynomials and are covered in detail in the module Polynomials.
Note that this is equal to the discriminant of the quadratic, so that if the roots are equal, the discriminant is 0. Newton developed the theory of symmetric functions and introduced the so called Newton identities that arise in higher algebra. These apply to the roots of polynomials. While the quadratic equation and the parabola were known from the days of the Greeks, higher order curves were not studied in depth until the calculus was developed.
Its graph is shown below. Focus-directrix definition of a parabola. There is also a geometric definition of the parabola in terms of the path traced out by a moving point. Such a graph is known as a locus. Suppose a is a positive real number. Now find all the points in the plane whose distance from S is equal to its perpendicular distance from the line d. Clearly the origin is one of those points. The point S is called the focus of the parabola and the line d is called the directrix.
Consider the vertex form of a parabola. Substitute the values of and into the formula. Cancel the common factor of and. Factor out of. Cancel the common factors. Cancel the common factor. Rewrite the expression. Divide by. Find the value of using the formula. Simplify each term.
Raising to any positive power yields. Multiply by. Substitute the values of , , and into the vertex form. Set equal to the new right side. Use the vertex form, , to determine the values of , , and. Since the value of is positive, the parabola opens up. Find the vertex.
Find , the distance from the vertex to the focus. Find the distance from the vertex to a focus of the parabola by using the following formula. Substitute the value of into the formula. Cancel the common factor of. Find the focus. The focus of a parabola can be found by adding to the y-coordinate if the parabola opens up or down. Substitute the known values of , , and into the formula and simplify.
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