Please answer this ASAP. The question is down below. Thank you!

Answer:
[tex]y = .5x^2 -2x -5[/tex]
Step-by-step explanation:
Well we can start by seeing if the parabola is the same width by comparing it to its parent function ( y = x^2 )
In y = x^2 the 2nd lowest point is just up 1 and right 1 away from the vertex.
This is not true for our parabola.
So we can widen it by to the desidered width by making the x^2 into a .5x^2.
So far we’ve got y = .5x^2
Now the parabola y intercept is at -5.
So we can add a -5 into the equation making it.
y = .5x^2 - 5
Now for the x value.
So we can find the x value by seeing how far away the parabola is from from the y axis.
So the x value is -2x.
So the full equation is [tex]y = .5x^2 -2x -5[/tex]
Look at the image below to compare.
Answer: [tex]Vertex: y=\dfrac{1}{2}(x-2)^2-7[/tex]
[tex]Standard: y=\dfrac{1}{2}x^2-2x-5[/tex]
Transformations: vertical shrink by a factor of 1/2,
horizontal shift 2 units to the right,
vertical shift 7 units down.
Step-by-step explanation:
Vertex form: y = a(x - h)² + k
Standard form: y = ax² + bx + c
Given: Vertex (h, k) = (2, -7), the y-intercept (0, c) = (0, -5)
Input those values into the Vertex form to solve for the a-value
-5=a(0 - 2)² - 7
2 = a(- 2)²
2 = 4a
[tex]\dfrac{1}{2}=a[/tex]
a) Input a = 1/2 and (h, k) = (2, -7) into the Vertex form
[tex]\large\boxed{y=\dfrac{1}{2}(x-2)^2-7}[/tex]
b) You can plug in a = 1/2, c = -5, (x, y) = (2, -7) to solve for "b"
or
You can expand the Vertex form (which is what I am going to do):
[tex]y=\dfrac{1}{2}(x-2)^2-7\\\\\\y=\dfrac{1}{2}(x^2-4x+4)-7\\\\\\y=\dfrac{1}{2}x^2-2x+2-7\\\\\\\large\boxed{y=\dfrac{1}{2}x^2-2x-5}[/tex]
c) Use the Vertex form to describe the transformations as follows:
[tex]y=\dfrac{1}{2}(x-2)^2-7[/tex]
a = 1/2 --> vertical shrink by a factor of 1/2
h = 2 --> horizontal shift 2 units to the right
k = -7 --> vertical shift 7 units down