Question

Consider the equation of parabola y = 5x - 30x + 45.

Its vertex is located at
( , )

Answer 1

Vertex is (3,0)

Explanation:

Given y= 5x^2 - 30x + 45

y = 5 ( x^2 - 6x + 9)

= 5 ( x -3 )^ 2 + 0

Standard formula of parabola is y = a(x - k)^2 + k where vertex is (h,k)

Here, h = 3 and k = 0

So, Vertex is ( 3, 0)

Answer 2

(3, 0 )

Explanation:

Given a parabola in standard form

y = ax² + bx + c (a ≠ 0 )

Then the x- coordinate of the vertex is

`x_(vertex)` = -
`(b)/(2a)`

y = 5x² - 30x + 45 ← is in standard form

with a = 5, b = - 30 , then

`x_(vertex)` = -
`(-30)/(10)` = 3

Substitute x = 3 into the function for corresponding value of y

y = 5(3)² - 30(3) + 45 = 45 - 90 + 45 = 0

vertex = (3, 0 )