Identify the vertex, axis of symmetry, and direction of opening for y=2(x+3)^2-5.
vertex = 1/2
axis = 1/2
direction = up
your text should have the following:
for y = a(x-p)^2 + q
vertex is (p,q)
axis of symmetry: x=p
opening: up if a>0, down if a<0
please redo your answers and let me know what you got.
y=2x+3^2-5
y=2x+9-5
x=-b/2a
x=-9/2(-9)
x=1/2
I don't know if I have the value for a right
vertex= (3,-5)
axis of symmetry= x=2
direction= up b/c 2>0
To identify the vertex, axis of symmetry, and direction of opening for the given equation y=2(x+3)^2-5, we can use the vertex form of a quadratic equation: y = a(x-h)^2 + k.
Comparing the given equation with the vertex form, we can determine the values of h (the x-coordinate of the vertex) and k (the y-coordinate of the vertex).
In this case, we have a = 2, h = -3, and k = -5.
To find the x-coordinate of the vertex, which is the axis of symmetry, we use the formula x = -b/2a. Since there is no x term (b = 0), the formula simplifies to x = -0/(2*2) = 0.
Therefore, the axis of symmetry is x = 0.
Substituting the x-coordinate of the vertex (h = -3) into the equation, we can determine the y-coordinate of the vertex: y = 2(-3+3)^2-5 = 2(0)^2-5 = 0-5 = -5.
Thus, the vertex is (0, -5).
Since the coefficient of the x^2 term (a) is positive (2 > 0), the parabola opens upward. Therefore, the direction of opening is up.