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Precalculus

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Help with these questions
help with these questions~

Thank you

1. Determine the interval(s) on which x^2 + 2x -3>0

a)x<-3, x>1
b)-3<x<1
c)x<-3, -3<x<1, x>1
d) x>1

Answer is D, x>1
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2. Determine when the function f(x)= 3x^3 + 4x^2 -59x -13 is greater than 7.

a) -5<x<1/3
b) -5<x<-1/3, x>4
c) x<4
d) x=-5, -1/3,4

Answer is: B
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3. Provde the intervals you would check to determine when -5x^2 + 37x> -15x^2 + 12x + 15.

a) x=3, x=-0.5
b) x<-3, -3<x<0.5, x>0.5
c) x<-1, -1<x<15, x>15
d) x= 1, x=15
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  • Precalculus - ,

    1. x^2 + 2x - 3 > 0
    (x+3)(x-1) > 0
    x-intercepts are -3 and 1
    so you are looking for the values of x when the parabola y = x^2 + 2x - 3 is above the x-axis
    Since it opens upwards those values are
    x < -3 OR x > 1
    The closest choice to that is #1, but they did not include the necessary OR. The comma is this context means AND, which would be incorrect.

    2. 3x^3 + 4x^2 - 59x - 13 > 7
    3x^3 + 4x^2 - 59x - 20 > 0
    after some quick tries of ±1, ±2, ± 4, I found x=4 to be a solution, so x-4 is a factor
    by synthetic division,
    3x^3 + 4x^2 - 59x - 20 = (x-4)(3x^2 + 16x + 5)
    = (x-4)(x+5)(3x+1)

    so the critical values are -5, -1/3, and 4
    So the curve is above the x-axis (or above 7 in the original) for
    -5 < x < -1/3 OR x > 4
    Again, they made an error by not stating the OR condition, but it looks like they meant b) , which is what you had.

    3. -5x^2 + 37x> -15x^2 + 12x + 15
    10x^2 + 25x - 15 > 0
    2x^2 + 5x - 3 > 0
    (2x - 1)(x + 3) > 0
    critical values are 1/s and -3
    so I would check
    x < -3, -3 < x < 1/2, and x > 1/2
    looks like b) is the correct choice.

  • Precalculus - ,

    1. The graph is a parabola, opening upward. So, there will be a left side and a right side above the x-axis. These intervals will be outside the two roots.

    A is the correct answer

    2. B is correct

    3. The graphs intersect at x = -3 and 0.5
    So, B is the answer

  • Precalculus - ,

    Thank you~

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