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Identify the interval(s) on which each quadratic function is positive.

1. y= x^2 + 2x - 8

2. y= -x^2 +4x +12

3. y= 5x^2 - 3x - 8

Help on these three problems above would be greatly appreciated, thanks! ☜(ˆ▿ˆc)

User Damianb
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1 Answer

4 votes

Answer:


\textsf{1. \quad $x < -4$ or $x > 2$}


\textsf{2. \quad $-2 < x < 6$}


\textsf{3. \quad $x < -1$ or $x > \frac{8}[5}$}
\textsf{3. \quad $x < -1$ or $x > \frac{8}[5}$}
\textsf{3. \quad $x < -1$ or $x > (8)/(5)$}

Explanation:

The intervals on which a quadratic function is positive are those intervals where the function is above the x-axis, i.e. where y > 0.

The zeros of the quadratic function are the points at which the parabola crosses the x-axis.

If the leading coefficient of a quadratic function is positive, the parabola opens upwards.

If the leading coefficient of a quadratic function is negative, the parabola opens downwards.

Therefore, to find the intervals on which each quadratic function is positive:

  • Calculate the zeros.
  • Determine if the parabola opens upwards or downwards.
  • If the parabola opens upwards, the intervals are less than the smaller zero and greater than the larger zero.
  • If the parabola opens downwards, the interval is between the zeros.

Question 1

Given function:


y=x^2+2x-8

Factor the given function:


\implies y= x^2+4x-2x-8


\implies y=x(x+4)-2(x+4)


\implies y=(x-2)(x+4)

Substitute y = 0 to find the zeros:


\implies (x-2)(x+4)=0


\implies x-2=0 \implies x=2


\implies x+4=0 \implies x=-4

The leading coefficient is positive, so the parabola opens upwards.

Therefore, the interval on which the function is positive is:

  • Solution: x < -4 or x > 2
  • Interval notation: (-∞, -4) ∪ (2, ∞)

Question 2

Given function:


y=-x^2+4x+12

Factor the given function:


\implies y=-(x^2-4x-12)


\implies y=-(x^2-6x+2x-12)


\implies y=-(x(x-6)+2(x-6))


\implies y=-(x+2)(x-6)

Substitute y = 0 to find the zeros:


\implies -(x+2)(x-6)=0


\implies (x+2)(x-6)=0


\implies x+2=0 \implies x=-2


\implies x-6=0 \implies x=6

The leading coefficient is negative, so the parabola opens downwards.

Therefore, the interval on which the function is positive is:

  • Solution: -2 < x < 6
  • Interval notation: (-2, 6)

Question 3

Given function:


y=5x^2-3x-8

Factor the given function:


\implies y=5x^2-8x+5x-8


\implies y=x(5x-8)+1(5x-8)


\implies y=(x+1)(5x-8)

Substitute y = 0 to find the zeros:


\implies (x+1)(5x-8)=0


\implies x+1=0 \implies x=-1


\implies 5x-8=0 \implies x=(8)/(5)

The leading coefficient is positive, so the parabola opens upwards.

Therefore, the interval on which the function is positive is:

  • Solution: x < -1 or x > ⁸/₅
  • Interval notation: (-∞, -1) ∪ (⁸/₅, ∞)
User Vishesh Chandra
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