Question

Dentify on which quadratic function is positive.

Y = 2x^2 - 17x + 30

Identify on which quadratic function is negative.

Y = - x^2 - 6x - 8

A explanation on the answers would be appreciated!

(Lots of points!)

Answer 1

Explanation:

Let us identify which quadratic function is positive. Yeah, let's start.

Y =
`{ \red{ \sf{2 {x}^(2) - 17x + 30}}}`

By using factorisation method,

`{ \red{ \sf{2 {x}^(2) - 12x - 5x + 30}}}`

Take common factors

`{ \red{ \sf{2x(x - 6) - 5(x - 6)}}}`

`{ \red{ \sf{(2x - 5)}}} \: \: \: \: \: \: \: || \: \: \: \: \: { \red{ \sf{(x - 6)}}}`

`{ \red{ \sf{2x - 5 = 0}}} \: \: || \: \: { \red{ \sf{x - 6 = 0}}}`

`{ \red{ \sf{2x = 5}}} \: \: \: \: \: \: \: \: \: || \: \: \: \: { \red{ \boxed{ \green{ \sf{x = 6}}}}}`

`{ \red{ \sf{{ ( \cancel2)/( \cancel2)x}}}} = { \red{ \sf{ (5)/(2)}}}`

`{ \red{ \boxed{ \green{ \sf{x = (5)/(2)}}}}}`

____________________________________

Y =
`{ \blue{ \sf {{ - x}^(2) - 6x - 8}}}`

By using factorisation method,

`{ \blue{ \sf{ - {x}^(2) - 2x - 4x - 8}}}`

Take common factors

`{ \blue{ \sf{ - x(x + 2) - 4(x + 2)}}}`

`{ \blue{ \sf{( - x - 4)}}} \: \: \: \: \: || \: \: \: \: \: { \blue{ \sf{(x + 2)}}}`

`{ \blue{ \sf{- x - 4 = 0}}} \: \: \: \: \: || \: \: \: \: \: { \blue{ \sf{x + 2 = 0}}}`

`{ \blue{ \boxed{ \green{ \sf{x = -4}}}}} \: \: \: \: \: || \: \: \: \: \: { \blue{ \boxed{ \green{ \sf{x = -2}}}}}`

Hence, the first quadratic function is positive and second quadratic function is negative.

Answer 2

`\textsf{$y = 2x^2 - 17x + 30$: \quad $\left(-\infty, (5)/(2)\right) \cup (6, \infty)$}`

`\textsf{$y = - x^2 - 6x - 8$: \quad $\left(-\infty, -4\right) \cup (-2, \infty)$}`

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:

`y = 2x^2 - 17x + 30`

Factor the equation:

`\implies y = 2x^2 - 17x + 30`

`\implies y = 2x^2 - 5x-12x + 30`

`\implies y=x(2x-5)-6(2x-5)`

`\implies y=(x-6)(2x-5)`

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:

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

`\implies 2x-5=0 \implies x=(5)/(2)`

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:

• 
  `\textsf{Solution: \quad $x < (5)/(2)$ \;and \;$x > 6$}`

• 
  `\textsf{Interval notation: \quad $\left(-\infty, (5)/(2)\right) \cup (6, \infty)$}`

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Given quadratic equation:

`y = - x^2 - 6x - 8`

Factor the equation:

`\implies y = - x^2 - 6x - 8`

`\implies y = -(x^2 +6x +8)`

`\implies y = -(x^2 +4x +2x+8)`

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

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

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:

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

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

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:

• 
  `\textsf{Solution: \quad $x < -4$ \;and \;$x > -2$}`

• 
  `\textsf{Interval notation: \quad $\left(-\infty, -4\right) \cup (-2, \infty)$}`