What is the vertex form of #y=4x^2-17x+60 #?

1 Answer

Answer:

Complete the square:
The vertex is #V_y(color(red)(17/8), color(red)(671/16))#

Explanation:

We can convert by completing the square on the first two terms, but first we need to have a "1" in front of the x-squared.

A standard form of parabola is:
#f(x)=ax^2+bx+c#
The vertex form for the same equation is:
#f(x)=a(x-color(red)h) + color(red)k#
Where the point #V(color(red)h,color(red)k)# is the vertex f(x)

#y=4(x^2-17/4x)+60#

Add (b/2)^2 to complete the square

#y=4(x^2-17/4x+289/64)+60-289/16#

The -289/16 is necessary to balance out the 4(289/64) that we added.

Factor the parentheses and find an LCD to add the 60 and -289/16

#y=4(x-17/8)^2+960/16-289/16#

#y=4(x-17/8)^2+671/16#