Answer:
[tex]\textsf{1. \quad $x < -4$ or $x > 2$}[/tex]
[tex]\textsf{2. \quad $-2 < x < 6$}[/tex]
[tex]\textsf{3. \quad $x < -1$ or $x > \dfrac{8}[5}$}[/tex][tex]\textsf{3. \quad $x < -1$ or $x > \dfrac{8}[5}$}[/tex][tex]\textsf{3. \quad $x < -1$ or $x > \dfrac{8}{5}$}[/tex]
Step-by-step 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:
[tex]y=x^2+2x-8[/tex]
Factor the given function:
[tex]\implies y= x^2+4x-2x-8[/tex]
[tex]\implies y=x(x+4)-2(x+4)[/tex]
[tex]\implies y=(x-2)(x+4)[/tex]
Substitute y = 0 to find the zeros:
[tex]\implies (x-2)(x+4)=0[/tex]
[tex]\implies x-2=0 \implies x=2[/tex]
[tex]\implies x+4=0 \implies x=-4[/tex]
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:
[tex]y=-x^2+4x+12[/tex]
Factor the given function:
[tex]\implies y=-(x^2-4x-12)[/tex]
[tex]\implies y=-(x^2-6x+2x-12)[/tex]
[tex]\implies y=-(x(x-6)+2(x-6))[/tex]
[tex]\implies y=-(x+2)(x-6)[/tex]
Substitute y = 0 to find the zeros:
[tex]\implies -(x+2)(x-6)=0[/tex]
[tex]\implies (x+2)(x-6)=0[/tex]
[tex]\implies x+2=0 \implies x=-2[/tex]
[tex]\implies x-6=0 \implies x=6[/tex]
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:
[tex]y=5x^2-3x-8[/tex]
Factor the given function:
[tex]\implies y=5x^2-8x+5x-8[/tex]
[tex]\implies y=x(5x-8)+1(5x-8)[/tex]
[tex]\implies y=(x+1)(5x-8)[/tex]
Substitute y = 0 to find the zeros:
[tex]\implies (x+1)(5x-8)=0[/tex]
[tex]\implies x+1=0 \implies x=-1[/tex]
[tex]\implies 5x-8=0 \implies x=\dfrac{8}{5}[/tex]
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, ∞)
