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anonymous

  • one year ago

Algebruh.

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  1. anonymous
    • one year ago
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    1.) Describe the direction of the parabola and determine the y-intercept and zeros of f(x) = 5x2 + 2x + 1 Find the vertex and the equation for the axis of symmetry of the parabola, showing your work, so Ray can include it in his coaster plan.

  2. anonymous
    • one year ago
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    @satellite73 @sammixboo @Mertsj @NeonStrawsForever @ParthKohli @e.mccormick

  3. anonymous
    • one year ago
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    leading coefficient is positive ( it is 5 ) so it opens up

  4. anonymous
    • one year ago
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    So far I have The vertex is going straight down, it is an acute angle. The y-intercept is 1

  5. anonymous
    • one year ago
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    the first coordinate of the vertex of \(y=ax^2+bx+c\) is \(-\frac{b}{2a}\) in your case it is \[-\frac{2}{2\times 5}=-\frac{1}{5}\]

  6. anonymous
    • one year ago
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    that makes the axis of symmetry \[x=-\frac{1}{5}\]

  7. anonymous
    • one year ago
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    the vertex is not an angle, acute or otherwise

  8. anonymous
    • one year ago
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    the vertex is a point the first coordinate of that point is \(-\frac{1}{5}\)

  9. anonymous
    • one year ago
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    the second coordinate of the vertex is what you get for \(y\) when you replace \(x\) by \(-\frac{1}{5}\)

  10. anonymous
    • one year ago
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    I meant parabola.

  11. anonymous
    • one year ago
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    Okay, here's what I have so far. 1.) Describe the direction of the parabola and determine the y-intercept and zeros of f(x) = 5x2 + 2x + 1 The parabola is going straight down, it is an acute angle, the y-intercept is 1.

  12. anonymous
    • one year ago
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    I need to find the zeroes of f(x) = 5x2 + 2x + 1

  13. anonymous
    • one year ago
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    @satellite73 @sammixboo @Empty @welshfella

  14. anonymous
    • one year ago
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    @DarkMoonZ @DaBest21

  15. welshfella
    • one year ago
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    well I can tell you there are no real zeroes to this

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