Identify the vertex, axis of symmetry, and direction of opening for y=2(x+3)^2-5
y = |2x+3|^2-5
h = 3
k = -5
vertex = (3,-5)
axis of symmetry = x = 3
direction = up b/c a = 2 and 2>0
y = 2 (x+3)^2 - 5
(x+3)^2 = (1/2)(y+5)
(x+3)^2 = 4 (1/8) (y+5)
vertex (-3,-5)
symmetric about x = -3
indeed opens up because a = +1/8, check as x becomes very positive or very negative, y gets big positive
I don't understand what you did after the 1st line.
Hmmm
I know, and I bet it is in your book, that
a formula of type:
(x-h)^2 = 4 a (y-k)
is a parabola with vertex at (h,k)
a is the distance from vertex to focus and from vertex to directrix
if a is +, opens up. if a is -, opens down
To find the vertex, axis of symmetry, and direction of opening for the given quadratic equation in the form y = a(x-h)^2 + k, where a, h, and k are the parameters, follow these steps:
1. Compare the given equation to the standard form: y = a(x-h)^2 + k.
2. Identify the values of h and k from the equation. In this case, h = -3 and k = -5.
3. The value of h represents the x-coordinate of the vertex, so the vertex of the parabola is (h, k). Therefore, the vertex is (3, -5).
4. The axis of symmetry is a vertical line passing through the vertex. The equation of the axis of symmetry is x = h. So, the axis of symmetry for this parabola is x = 3.
5. To determine the direction of opening, analyze the coefficient "a." If a is positive, the parabola opens upwards, and if a is negative, it opens downwards. In this case, a = 2, which is positive, so the parabola opens upward.
Therefore, for the equation y = 2(x+3)^2-5, the vertex is (3, -5), the axis of symmetry is x = 3, and the direction of opening is upward.