Answer:
[tex]\textsf{1. \quad Solution: $1 < x < 4$,\quad Interval notation: $(1, 4)$}[/tex]
[tex]\textsf{2. \quad Solution: $-2 < x < 4$,\quad Interval notation: $(-2, 4)$}[/tex]
Step-by-step explanation:
Question 1
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.
As the given quadratic function has a negative leading coefficient, the parabola opens downwards. Therefore, the interval on which y > 0 is between the zeros.
To find the zeros of the given quadratic function, substitute y = 0 and factor:
[tex]\begin{aligned}y&= 0\\\implies -7x^2+35x-28& = 0\\-7(x^2-5x+4)& = 0\\x^2-5x+4& = 0\\x^2-x-4x+4& = 0\\x(x-1)-4(x-1)&= 0\\(x-1)(x-4)& = 0\end{aligned}[/tex]
Apply the zero-product property:
[tex]\implies x-1=0 \implies x=1[/tex]
[tex]\implies x-4=0 \implies x=4[/tex]
Therefore, the interval on which the function is positive is:
- Solution: 1 < x < 4
- Interval notation: (1, 4)
Question 2
The intervals on which a quadratic function is negative are those intervals where the function is below 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.
As the given quadratic function has a positive leading coefficient, the parabola opens upwards. Therefore, the interval on which y < 0 is between the zeros.
To find the zeros of the given quadratic function, substitute y = 0 and factor:
[tex]\begin{aligned}y&= 0\\\implies 2x^2-4x-16& = 0\\2(x^2-2x-8)& = 0\\x^2-2x-8& = 0\\x^2-4x+2-8& = 0\\x(x-4)+2(x-4)&= 0\\(x+2)(x-4)& = 0\end{aligned}[/tex]
Apply the zero-product property:
[tex]\implies x+2=0 \implies x=-2[/tex]
[tex]\implies x-4=0 \implies x=4[/tex]
Therefore, the interval on which the function is negative is:
- Solution: -2 < x < 4
- Interval notation: (-2, 4)
