What is the vertex form of y=17x^2+88x+1?

1 Answer
Nallasivam V
Jul 30, 2017

y=17(x+44/17)-1919/17

Explanation:

Given -

y=17x^2+88x+1

Vertex
x-coordinate of the vertex

x=(-b)/(2a)=(-88)/(2xx 17)=(-88)/34=(-44)/17

y-coordinate of the vertex

y=17((-44)/17)^2+88((-44)/17)+1
y=17((1936)/289)-3872/17+1
y=32912/289-3872/17+1
y=(32912-65824+289)/289=(-32623)/289=(-1919)/17

The vertex form of the equation is

y=a(x-h)^2+k

a=17 coefficient of x^2
h=(-44)/17 x coordinate of the vertex
k=(-1919)/17 y-coordinate of the vertex

y=17(x+44/17)-1919/17

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