Question

can someone please help? :((
if you want another 50 points or more you could maybe help with one of the previous unanswered questions on my profile

Answer 1

Just find the vertex and then compare with graph

#a

• y=2x²-8
• y=2(x-0)²-8

Parabola opening upwards

• Vertex at (0,-8)

Graph 3

#2

• y=(x+3)²+0

Vertex at (-3,0)

Graph IV

#3

• y=-2(x-4)²+8

Parabola opening downwards as a is -ve

Graph I

#4

One graph is left

• Graph Ii

Answer 2

Answer:

a)  graph iii)

b)  graph iv)

c)  graph i)

d)  graph ii)

Step-by-step explanation:

Vertex form of a quadratic equation:  $y = a(x - h)^2 + k$

where $(h, k)$ is the vertex (turning point)

First, determine the vertices of the parabolas by inspection of the graphs:

• Graph i) → vertex = (4, 8)
• Graph ii) → vertex = (3, -8)
• Graph iii) → vertex = (0, -8)
• Graph iv) → vertex = (-3, 0)

Next, write each given equation in vertex form and compare to the vertices above.

$\textsf{a)}\quad y=2x^2-8$

$\textsf{Vertex form}: \quad y=2(x-0)^2-8$

⇒ Vertex = (0, -8)

Therefore, graph iii)

$\textsf{b)} \quad y=(x+3)^2$

$\textsf{Vertex form}: \quad y=(x+3)^2+0$

⇒ Vertex = (-3, 0)

Therefore, graph iv)

$\textsf{c)} \quad y=-2|x-4|^2+8$

$\textsf{Vertex form}: \quad y=-2|x-4|^2+8$

⇒ Vertex = (4, 8)

Therefore, graph i)

$\textsf{d)} \quad y=(x-3)^2-8$

$\textsf{Vertex form}: \quad y=(x-3)^2-8$

⇒ Vertex = (3, -8)

Therefore, graph ii)