Use this form: y= a(x-h)² + k ; plug in the x and y coordinates of the vertex into (h,k) and then the other point coordinates into (x,y) and solve for a.
The vertex of a parabola is found by using the solution of the equation -b/2a and putting it into the quadratic equation. a is the coefficient of x^2. b is the coefficient of the other x in the equation. Ex. y=2x^2+2x+1 -b/2a = -2/2(2) = -1/2 Now put -1/2 in the place of every x in the equation. y=2(-1/2)^2+2(-1/2)+1 The vertex is (-1/2, 1/2)
False, other geometric objects exist which can be defined as a parrticular locus of points, such as the parabola and the hyperbola.
Two adjacent sides of a polygon will meet at a vertex. End points of any other pairs of sides do not have a specific name.
cube has more than one vertex and many other shapes i hope this helps you answer your questionthanks,Chadai age 11
The vertex of a parabola doe not provide enough information to graph anything - other than the vertex!
Parabola's equations can be any second-order polynomial, but popular forms are: Standard form for a parabola is y=Ax2+Bx+C This is a very nice form for algebraic use, but physically, A B and C do not directly represent the shape of the curve Vertex form: y=a(x-h)2+k Often people use this because (h, k) would also be the location of the parabola's vertex. The a at the beginning represents the "wideness" (tight parabola's have high a values greater than one; wide parabola's have low a values less than one) and the direction of the parabola (negative a means it opens downward while a positive a mean it opens upwards) Factored form: y=a(x-r)(y-s) This form is very nice because it is factored, and it shows you the zeros of the function: r and s. a is the same as from the vertex form. This is less popular than other forms because some parabolas have imaginary numbers for r and s when in factored form, even if the parabola is made of completely real points.
There is not enough information. You need either the directrix or vertex (or some other item of information).
Quadratic equations always have 2 solutions. The solutions may be 2 real numbers (think of a parabola crossing the x axis at 2 different points) or it could have a "double root" real solution (think of a parabola just touching the x-axis at its vertex), or it can have complex roots (which will be complex conjugates of each other). For the last scenario, the graph of the parabola will not touch the x axis.
Use this form: y= a(x-h)² + k ; plug in the x and y coordinates of the vertex into (h,k) and then the other point coordinates into (x,y) and solve for a.
Geometry.
By the geometric definition of a line, it is represented by two points, and all points on the line are collinear, between or extrapolating to infinity from the straight line made by the two points. In other words, a line is straight, and can be represented by a binomial function (example: y=2x+1). A parabola is a function, but cannot be described mathematically as a line.
The vertex of a parabola is found by using the solution of the equation -b/2a and putting it into the quadratic equation. a is the coefficient of x^2. b is the coefficient of the other x in the equation. Ex. y=2x^2+2x+1 -b/2a = -2/2(2) = -1/2 Now put -1/2 in the place of every x in the equation. y=2(-1/2)^2+2(-1/2)+1 The vertex is (-1/2, 1/2)
The graph of a quadratic function is y = ax + bx + c or y = a(x - h) + k, where (h, k) is the vertex. Since x2 - 12x + 40 cannot be factored, then I will complete the square which will give me the vertex. So I will draw the parabola (which opens upward) by using the vertex and the y-intercept, (0, 40). y = x2 - 12x + 40 add and subtract 62 y = x2 - 12x + 62 - 36 + 40 y = (x - 6)2 + 4 so the vertex is (6, 4) which shows that the parabola does not intersect the x-axis. Since the axis of symmetry is x = 6, and the y-intercept point is (0, 40), I have another point (12, 40). I can also find other points by letting x to be 2, or 3, or 4 and finding the corresponding y-values. For example, if x = 3, then y = 13. So the point (3, 13) will give me its symmetric point (9, 13), and so on. Thus, I can use all these points to draw the parabola.
-3x2+12x-1 = y Solve the quadratic equation when y = 0 by means of the quadratic equation formula which gives x values of 3.914854216 and 0.085145784. Add these values together and divide them by 2 which is 4/2 = 2 and this is the line symmetry of the parabola. Substitute 2 for x into the original equation to find the value of y: So the vertex is at (2, 11) Remember that the parabola has a maximum value because the coefficient of x2 is negative in other words it will face downwards.
The point of a triangle is called a vertex, same goes with rectangles and all other shapes with points.
For example, y = ax2 + bx + c (the equation of a parabola). Every parabola has an axis of symmetry and the graph to either side of the axis of symmetry is a mirror image of the other side. It means that if we know a point on one side of the parabola, we can find its symmetric point to the other side, based on the axis of symmetry. Those symmetric points have opposite x-coordinate values, and the same y-coordinate value. The vertex only is a single point which lies on the axis of symmetry.