Question

NEED MATH HELP NOW!!!PLEASE HELP!!!

Answer 1

Answer:

Step-by-step explanation:

The given equation of a parabola is,

$y=-\frac{1}{2}(x+3)^2+4$

When we compare this equation with the vertex form of the quadratic equation,

y = a(x - h)² + k

where (h, k) is the vertex,

Vertex of the parabola,

(-3, 4)

y-intercept,

$y=-\frac{1}{2}(0+3)^2+4$

$y=-\frac{9}{2}+4$

$y=-\frac{1}{2}$

x-intercepts,

$0=-\frac{1}{2}(x+3)^2+4$

$\frac{1}{2}(x+3)^2=4$

(x + 3) = ±√8

x = (3 - 2√2), (3 + 2√2)

Axis of symmetry,

x = -3

Domain:

(-∞, ∞)

Range:  

(-∞, 4]