Question

Dentify on which quadratic function is positive.

Answer 1

Answer:

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

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

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:

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

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:

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

$\implies 2x-5=0 \implies x=\dfrac{5}{2}$

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:

• $\textsf{Solution: \quad x < \dfrac{5}{2} \;and \; x > 6 }$

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

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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:

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

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

Answer 2

Step-by-step 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

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

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

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

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

${ \red{ \sf{{ \frac{ \cancel2}{ \cancel2}x}}}} = { \red{ \sf{ \frac{5}{2}}}}$

${ \red{ \boxed{ \green{ \sf{x = \frac{5}{2}}}}}}$

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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.