What is the vertex form of $y = 4 x^{2} - x + 4$ $y=4x^2-x+4$ ?

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Answer 1 · 2016-06-03T10:56:57

The vertex is at $(\frac{1}{8}, \frac{63}{16})$ $(1/8,63/16)$

Your quadratic equation is of the form

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

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

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

$y = 4 x^{2} - x + 4$ $y=4x^2-x+4$

$y = 4 x^{2} - x + \frac{4}{64} - \frac{4}{64} + 4$ $y=4x^2-x+ color(red)(4/64) - color(red)( 4/64)+4$

$y = (4 x^{2} - x + \frac{4}{64}) - \frac{4}{64} + 4$ $y=(4x^2-x+ color(red)(4/64) )- color(red)( 4/64)+4$

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

$y = 4 (x^{2} - \frac{1}{4} x + \frac{1}{64}) - \frac{4}{64} + 4$ $y=4(x^2-1/4x+ color(red)(1/64) )- color(red)( 4/64)+4$

$y = 4 {(x - \frac{1}{8})}^{2} + \frac{4 \times 64 - 4}{64}$ $y=4(x - 1/8 )^2 + (4xx64-4)/64$

$y = 4 {(x - \frac{1}{8})}^{2} + \frac{252}{64}$ $y=4(x - 1/8 )^2 + 252/64$

$y = 4 {(x - \frac{1}{8})}^{2} + \frac{63}{16}$ $y=4(x - 1/8 )^2 + 63/16$

The vertex is at $(\frac{1}{8}, \frac{63}{16})$ $(1/8,63/16)$

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

What is the vertex form of y=4x2−x+4y=4x^2-x+4?