Question

What is the vertex form of `y=4x^2+5x+2`?

Answer 1

`y = 4(x + 5/8)^2 + 7/16`

Explanation:

The standard form of the quadratic function is :`y = ax^2+bx+c`

The function: `y = 4x^2 + 5x + 2" is in this form "`

with a = 4 , b = 5 and c = 2
>`"--------------------------------------------------"`
The vertex form of the quadratic function is

`y = a(x - h )^2 + k" (h,k) are the coords of vertex "`

x-coord of vertex (h) `= -b/(2a) = -5/(2xx4) = - 5/8`
now substitute `x = -5/8 " into " y = 4x^2+5x+2`
y-coord of vertex (k) = `4(-5/8)^2 + 5(-5/8 )+ 2`
`= 4(25/64) - 25/8 + 2 = 7/16`
hence vertex has coordinates `(-5/8 , 7/16 )`
> `"------------------------------------------------"`
so a = 4 and (h , k ) `= (-5/8 , 7/16 )`

`rArr" vertex form is " y = 4(x + 5/8)^2 + 7/16`