What is the standard form of $y = (x - 8) (4 x + 1)$ $y= (x - 8) (4x+1)$ ?

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What is the standard form of $y = (x - 8) (4 x + 1)$ $y= (x - 8) (4x+1)$ ?

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Answer 1 · 2016-03-03T12:49:09

The quadratic form is y = $4 x^{2} - 31 x - 8$ $4x^2-31x-8$ .
If it is for making a graph, the standard form is
${(x - \frac{31}{8})}^{2} = 4 (\frac{1}{16}) (y + \frac{993}{16})$ $(x-31/8)^2 = 4(1/16)(y+993/16)$ .

Explanation:

The equation represents the parabola with .
Vertex: $(\frac{31}{8}, - \frac{993}{16})$ $(31/8, -993/16)$ ,
Axis: Parallel to +ve y-axis,
Focus at $(\frac{31}{8}, - \frac{993}{16} + \frac{1}{16})$ $(31/8, -993/16 +1/16)$ and
Directrix along x = $(\frac{31}{8}, - \frac{993}{16} - \frac{1}{16})$ $(31/8, -993/16 -1/16)$ .

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What is the standard form of $y = (x - 8) (4 x + 1)$ $y= (x - 8) (4x+1)$ ?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

What is the standard form of y= (x - 8) (4x+1) y=(x−8)(4x+1)y= (x - 8) (4x+1) ?

1 Answer

Explanation:

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What is the standard form of $y = (x - 8) (4 x + 1)$ $y= (x - 8) (4x+1)$ ?