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What is the equation of a parabola with the vertex of 2 -1?
3
A parabola has a vertex at -3 -2 what is its equation?
-1
What is the standard equation for vertex at origin opens down 1 and 76 units between the vertex and focus?
Since the vertex is at the origin and the parabola opens downward, the equation of the parabola is x2 = 4py, where p < 0, and the axis of symmetry is the y-axis. So the focus is at y-axis at (0, p) and the directrix equation is y = -p. Now, what do you mean with 1 and 76 units? 1.76 units? If the distance of the vertex and the focus is 1.76 units, then p = -1.76, thus 4p = -7.04, then the equation of the parabola is x2 = -7.04y.
What is the coefficient of the x² term of the parabola with a vertex at -3 -1 and passes through the point 4 0?
A parabola with vertex (h, k) has equation: y = a(x - h)² + k With vertex (-3, -1) this becomes: y = a(x - -3)² + -1 = a(x + 3)² - 1 The point (4, 0) is on this parabola so: 0 = a(4 + 3)² - 1 → 7²a = 1 → a = 1/49 Thus the coefficient of x² is 1/49.
Does a parabola always have roots and a vertex?
An x2 parabola will always have one vertex, but depending on the discriminant of the function (b2-4ac) the parabola will either have 2 roots (it crosses the x-axis twice), 1 repeating root (the parabola meets the x-axis at a single point), or no real roots (the parabola doesn't meet the x-axis at all)
The vertex of the parabola below is at the point 4 -1 which equation be this parabola's equation?
5
What is the equation of a parabola with the vertex of 2 -1?
3
The vertex of the parabola below is at the point -2 1 Which of the equations below could be this parabolas equation?
Go study
The vertex of this parabola is at 5 5 When the x-value is 6 the y-value is -1 What is the coefficient of the squared expression in the parabola's equation?
The vertex of this parabola is at 5 5 When the x-value is 6 the y-value is -1. The coefficient of the squared expression in the parabola's equation is -6.
A parabola has a vertex at -3 -2 what is its equation?
-1
What is the standard equation for vertex at origin opens down 1 and 76 units between the vertex and focus?
Since the vertex is at the origin and the parabola opens downward, the equation of the parabola is x2 = 4py, where p < 0, and the axis of symmetry is the y-axis. So the focus is at y-axis at (0, p) and the directrix equation is y = -p. Now, what do you mean with 1 and 76 units? 1.76 units? If the distance of the vertex and the focus is 1.76 units, then p = -1.76, thus 4p = -7.04, then the equation of the parabola is x2 = -7.04y.
The vertex of this parabola is at -3 -1 When the y-value is 0 the x-value is 4 What is the coefficient of the squared term in the parabolas equation?
The vertex of this parabola is at -3 -1 When the y-value is 0 the x-value is 4. The coefficient of the squared term in the parabolas equation is 7
What is the coefficient of the squared term in the parabola's equation when the vertex is at 2 -1 and the point 5 0 is on it?
A parabola with vertex (h, k) has equation of the form: y = a(x - h)² + k → vertex (k, h) = (2, -1), and a point on it is (5, 0) → 0 = a(5 - 2)² + -1 → 0 = a(3)² -1 → 1 = 9a → a = 1/9 → The coefficient of the x² term is 1/9
What is the coefficient of the squared term in the parabola's equation when the vertex is at -2 -3 and the point -1 -5 is on it?
A parabola with vertex (h, k) has equation of the form: y = a(x - h)² + k → vertex (k, h) = (-2, -3), and a point on it is (-1, -5) → -5 = a(-1 - -2)² + -3 → -5 = a(1)² - 3 → -5 = a - 3 → a = -2 → The coefficient of the x² term is -2.
What is the coefficient of the squared term in the parabola's equation when the vertex is at 3 5 and the point -1 6 is on it?
A parabola with vertex (h, k) has equation of the form: y = a(x - h)² + k → vertex (k, h) = (3, 5), and a point on it is (-1, 6) → 6 = a(-1 - 3)² + 5 → 6 = a(-4)² + 5 → 1 = 16a → a = 1/16 → The coefficient of the x² term is 1/16
What is the vertex of a parabola?
the vertex of a parabola is the 2 x-intercepts times-ed and then divided by two (if there is only 1 x-intercept then that is the vertex)
Vertex of a parabola?
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)
