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

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Answer 1 · 2016-06-03T10:56:57

The vertex is at #(1/8,63/16)#

Your quadratic equation is of the form

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

The vertex is at the point #(h,k)#

Rearrange your equation to obtain a form similar to that of the quadratic equation.

#y=4x^2-x+4#

#y=4x^2-x+ color(red)(4/64) - color(red)( 4/64)+4#

#y=(4x^2-x+ color(red)(4/64) )- color(red)( 4/64)+4#

Take #color(red)4# as a common factor.

#y=4(x^2-1/4x+ color(red)(1/64) )- color(red)( 4/64)+4#

#y=4(x - 1/8 )^2 + (4xx64-4)/64#

#y=4(x - 1/8 )^2 + 252/64#

#y=4(x - 1/8 )^2 + 63/16#

The vertex is at #(1/8,63/16)#

graph{4*x^2-x+4 [-7.8, 8.074, -1.044, 6.896]}