What is the vertex form of y= 8x^2+17x+1 y=8x2+17x+1?
1 Answer
Jan 26, 2016
y =8 (x + 17/16 )^2 - 257/32 y=8(x+1716)2−25732
Explanation:
The vertex form of the trinomial is;
y = a(x - h )^2 + ky=a(x−h)2+k where (h , k ) are the coordinates of the vertex.
the x-coordinate of the vertex is x
= -b/(2a)=−b2a [from
8x^2 + 17x + 1 8x2+17x+1 a = 8 , b = 17 and c = 1 ]
so x-coord
= -17/16 =−1716 and y-coord
= 8 xx (-17/16)^2 + 17 xx (-17/16) + 1 =8×(−1716)2+17×(−1716)+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
