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
When the coefficient of ( x^2 ) is negative in a quadratic equation, the parabola opens downward. This means that the vertex of the parabola represents a maximum point, and the value of the function decreases on either side of the vertex. Consequently, the graph will touch or cross the x-axis at most twice, indicating that the quadratic can have zero, one, or two real roots.
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.
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.
-5
-3
-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
To have a parabola with only one x-intercept, the vertex of the parabola must lie on the x-axis. This means the parabola opens either upwards or downwards, depending on the coefficient of the squared term in the equation. If the coefficient is positive, the parabola opens upwards, and if it is negative, the parabola opens downwards. By adjusting the coefficients in the equation of the parabola, you can position the vertex such that there is only one x-intercept.
No.
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, -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.
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