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.
No.
7
The vertex of a parabola doe not provide enough information to graph anything - other than the vertex!
The vertex is either the minimum (very bottom) or maximum (very top) of a parabola.
A median of a triangle is a line or segment that passes through a vertex and the midpoint of the side opposite that vertex. The median only bisects the vertex angle from which it is drawn when it is an isosceles triangle.
please help
The vertex of this parabola is at -2 -3 When the y-value is -2 the x-value is -5. The coefficient of the squared term in the parabola's equation is -3.
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.
-3
-5
-3
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
No.
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.
The vertex would be the point where both sides of the parabola meet.
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
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