Answer:
[tex]\textsf{$y = 2x^2 - 17x + 30$: \quad $\left(-\infty, \dfrac{5}{2}\right) \cup (6, \infty)$}[/tex]
[tex]\textsf{$y = - x^2 - 6x - 8$: \quad $\left(-\infty, -4\right) \cup (-2, \infty)$}[/tex]
Step-by-step explanation:
A function is positive when it is above the x-axis, and negative when it is below the x-axis.
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Given quadratic equation:
[tex]y = 2x^2 - 17x + 30[/tex]
Factor the equation:
[tex]\implies y = 2x^2 - 17x + 30[/tex]
[tex]\implies y = 2x^2 - 5x-12x + 30[/tex]
[tex]\implies y=x(2x-5)-6(2x-5)[/tex]
[tex]\implies y=(x-6)(2x-5)[/tex]
The x-intercepts of the parabola are when y = 0.
To find the x-intercepts, set each factor equal to zero and solve for x:
[tex]\implies x-6=0 \implies x=6[/tex]
[tex]\implies 2x-5=0 \implies x=\dfrac{5}{2}[/tex]
Therefore, the x-intercepts are x = ⁵/₂ and x = 6.
The leading coefficient of the given function is positive, so the parabola opens upwards.
The function is positive when it is above the x-axis.
Therefore, the function is positive for the values of x less than the smallest x-intercept and more than the largest x-intercept:
- [tex]\textsf{Solution: \quad $x < \dfrac{5}{2}$ \;and \;$x > 6$}[/tex]
- [tex]\textsf{Interval notation: \quad $\left(-\infty, \dfrac{5}{2}\right) \cup (6, \infty)$}[/tex]
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Given quadratic equation:
[tex]y = - x^2 - 6x - 8[/tex]
Factor the equation:
[tex]\implies y = - x^2 - 6x - 8[/tex]
[tex]\implies y = -(x^2 +6x +8)[/tex]
[tex]\implies y = -(x^2 +4x +2x+8)[/tex]
[tex]\implies y = -((x(x+4)+2(x+4))[/tex]
[tex]\implies y = -(x+4)(x+2)[/tex]
The x-intercepts of the parabola are when y = 0.
To find the x-intercepts, set each factor equal to zero and solve for x:
[tex]\implies x+4=0 \implies x=-4[/tex]
[tex]\implies x+2=0 \implies x=-2[/tex]
Therefore, the x-intercepts are x = -4 and x = -2.
The leading coefficient of the given function is negative, so the parabola opens downwards.
The function is negative when it is below the x-axis.
Therefore, the function is negative for the values of x less than the smallest x-intercept and more than the largest x-intercept:
- [tex]\textsf{Solution: \quad $x < -4$ \;and \;$x > -2$}[/tex]
- [tex]\textsf{Interval notation: \quad $\left(-\infty, -4\right) \cup (-2, \infty)$}[/tex]
