Question

Our parabola runs through (3,5) and also through (5,5), and it has of 2. What is the vertex form(s) of this quadratic equation? Draw a sketch if necessary to help you out.( |a|- is absolute value of a)

Answer 1

it has (BLANK) of 2..... what is (BLANK)?

Answer 2

the a value, sorry

Answer 3

Well our equation is

y=2(x-h) + k

Since a = 2, right?

Answer 4

yes

Answer 5

You can start with y = ax^2 + bx + c
where a = 2
You are given two points. So you can solve for b and c.

Then you can complete the square and put it in the vertex form.

Answer 6

y = 2x^2 + bx +c ---- (1)
(3,5) and (5,5) are points on the parabola.
when x = 3, y = 5. Substitute in (1)
5 = 2(3)^2 + 3b + c
5 = 18 + 3b + c
3b + c = -13 ----- (2)
when x = 5, y = 5. Substitute in (1)
5 = 2(5)^2 + 5b + c
5 = 50 + 5b + c
5b + c = -45 ------ (3)
Subtract (2) from (3)
2b = -45 + 13 = -32
b = -16
Put it in (2):
3(-16) + c = -13
c = -13 + 48 = 35

The parabola is: y = 2x^2 -16x + 35

Complete the square to put the equation in vertex form:
y = 2x^2 -16x + 35 = 2(x^2 - 8x + 35/2)
y = 2{ (x-4)^2 - 16 + 35/2 }
y = 2 { (x-4)^2 + (-32 + 35)/2 }
y = 2 { (x-4)^2 + (3)/2 }
y = 2(x-4)^2 + 3

Compare this to the standard equation of a parabola in vertex form: a(x-h)^2 + k
where (h,k) is the vertex and you can find the vertex of the parabola y = 2(x-4)^2 + 3

Answer 7

But vertex is not asked. Just the vertex form of the parabola.