Answer:
Equation: f(x) = 2(x + 5)^2 + 2
Vertex: (-5, 2)
Step-by-step explanation:
The form the question wants us to write the quadratic function in is called "vertex form":
f(x) = a (x - h)^2 + k
a = the a in a standard quadratic equation (y = ax^2 + bx + c) or the coefficient of the x^2
h = x coordinate of the vertex
k = y coordinate of the vertex
To find the vertex, we are going to use the quadratic equation given:
2x^2 + 20x + 52
Comparing it to the standard quadratic equation (y = ax^2 + bx + c),
a = 2
b = 20
c = 52
Now we can start finding our vertex.
To find h, we are going to use this formula:
-b / 2a
We already know b = 20 & a = 2, so we can just substitute that into our formula:
- (20) / 2*2
Which equals:
-20/4 = -5
So h (or the x coordinate of the vertex) is equal to -5
Next we will find k, or the y coordinate of the vertex.
To do that, we are going to plug in -5 into 2x^2 + 20x + 52:
2(-5)^2 + 20(-5) + 52
2(25) -100 + 52
50 - 100 + 52
-50 + 52
2
k (or the y coordinate of the vertex) is equal to 2
The vertex is (-5, 2)
However, we still need to find our equation in vertex form.
We know a = 2, h = -5, & k = 2. Now we substitute these into our vertex form equation:
a(x - h)^2 + k
(2)(x - (-5))^2 + (2)
2(x + 5)^2 + 2
(Remember that the -5 cancels with the - in front of it, making it a positive 5)
The equation is f(x) = 2(x + 5)^2 + 2
Hope it helps (●'◡'●)
