Question

What is the vertex form of `y= 8x^2+17x+1`?

Answer 1

`y =8 (x + 17/16 )^2 - 257/32`

Explanation:

The vertex form of the trinomial is; `y = a(x - h )^2 + k`

where (h , k ) are the coordinates of the vertex.

the x-coordinate of the vertex is x `= -b/(2a)`

[from `8x^2 + 17x + 1`

a = 8 , b = 17 and c = 1 ]

so x-coord`= -17/16`

and y-coord `= 8 xx (-17/16)^2 + 17 xx (-17/16) + 1`

`= cancel(8) xx 289/cancel(256) - 289/16 + 1`

`= 289/32 - 578/32 + 32/32 = -257/32`

Require a point to find a: if x = 0 then y=1 ie (0,1)

and so : 1= a`(17/16)^2 -257/32 = (289a)/256 -257/32`

hence `a = (256 + 2056)/289 = 8`

equation is : `y = 8( x + 17/16)^2 - 257/32`