Answer:
Its C boys
Step-by-step explanation:
I graphed all of them and compared
The equation that could represent the graph of the parabola is (c) [tex]y = -2x^2 - 16x- 28[/tex]
From the question, we have the following highlights
The second highlight above means that:
The x coordinate (h) of the vertex is negative, while the y-coordinate (k) is positive.
For a parabola: [tex]y = ax^2 + bx + c[/tex], the vertex is:
[tex](h,k) = (-\frac{b}{2a},f(h))[/tex]
Next, we test the options
(a) y = –2x2 + 3x – 5
We have:
[tex]h = -\frac{3}{-2 \times 2}[/tex]
[tex]h = 0.75[/tex]
The value of h is positive.
So, this cannot represent the parabola
(b) y = –2x2 – 4x – 2
We have:
[tex]h = -\frac{-4}{-2\times 2}[/tex]
[tex]h = -1[/tex]
Substitute -1 for x in the function
[tex]k = -2x^2 - 4x - 2[/tex]
[tex]k = -2(-1)^2 - 4(-1) - 2[/tex]
[tex]k = 0[/tex]
The value of k cannot be 0
So, this cannot represent the parabola
(c) y = –2x2 – 16x– 28
We have:
[tex]h = -\frac{-16}{-2\times 2}[/tex]
[tex]h = -4[/tex]
Substitute -4 for x in the function
[tex]k = -2x^2 - 16x- 28[/tex]
[tex]k = -2(-4)^2 - 16(-4)- 28[/tex]
[tex]k=4[/tex]
The x coordinate (h) of the vertex is negative, while the y-coordinate (k) is positive.
Hence, [tex]y = -2x^2 - 16x- 28[/tex] could represent the parabola
Read more about parabola at:
https://brainly.com/question/17987697
