Exploring Positive and Negative Domains: y = x² + 2x + 4.2

Quadratic Functions with Negative Discriminants

Find the positive and negative domains of the following function:

y=x2+2x+415 y=x^2+2x+4\frac{1}{5}

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Step-by-step written solution

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1

Understand the problem

Find the positive and negative domains of the following function:

y=x2+2x+415 y=x^2+2x+4\frac{1}{5}

2

Step-by-step solution

To find the positive and negative domains of the function y=x2+2x+4.2 y = x^2 + 2x + 4.2 , we need to consider the graph of this function and its roots.

First, let's compute the discriminant of the quadratic y=x2+2x+4.2 y = x^2 + 2x + 4.2 . The discriminant Δ \Delta is given by b24ac b^2 - 4ac .

Here, a=1 a = 1 , b=2 b = 2 , and c=4.2 c = 4.2 .

Calculating, we have:

Δ=22414.2=416.8=12.8 \Delta = 2^2 - 4 \cdot 1 \cdot 4.2 = 4 - 16.8 = -12.8 .

Since the discriminant is negative, there are no real roots. This means the parabola does not intersect the x-axis.

Next, because a=1 a = 1 is positive, the parabola opens upwards.

Hence, the entire parabola lies above the x-axis, indicating that the function y=x2+2x+4.2 y = x^2 + 2x + 4.2 is positive for all real x x .

Thus, there is no negative domain for this quadratic since it doesn't dip below the x-axis at any point.

Therefore, the positive and negative domains are:

x>0: x > 0 : for all x x

x<0: x < 0 : none

3

Final Answer

x>0: x > 0 : for all x x

x<0: x < 0 : none

Key Points to Remember

Essential concepts to master this topic
  • Discriminant Rule: When Δ < 0, parabola has no real roots
  • Technique: Calculate b24ac=416.8=12.8 b^2 - 4ac = 4 - 16.8 = -12.8 to determine sign
  • Check: Positive leading coefficient + negative discriminant = always positive function ✓

Common Mistakes

Avoid these frequent errors
  • Confusing positive/negative domains with positive/negative x-values
    Don't think 'positive domain' means x > 0 = wrong interpretation! This confuses the sign of x with the sign of y. Always remember: positive domain means where y > 0, not where x > 0.

Practice Quiz

Test your knowledge with interactive questions

The graph of the function below intersects the X-axis at points A and B.

The vertex of the parabola is marked at point C.

Find all values of \( x \) where \( f\left(x\right) > 0 \).

AAABBBCCCX

FAQ

Everything you need to know about this question

What does 'positive domain' actually mean?

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The positive domain is where the function output (y-values) is positive, not where x is positive! For y=x2+2x+4.2 y = x^2 + 2x + 4.2 , this function is positive for all real numbers.

Why don't I need to find the roots first?

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When the discriminant is negative, there are no real roots! The parabola never touches the x-axis, so you can determine the sign just from the discriminant and leading coefficient.

How do I know the parabola opens upward?

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Look at the leading coefficient (the coefficient of x2 x^2 ). Since a=1>0 a = 1 > 0 , the parabola opens upward. If a were negative, it would open downward.

What if the discriminant were positive instead?

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With a positive discriminant, you'd have two real roots. The function would be negative between the roots and positive outside them. Always find the roots first in that case!

Can a quadratic function be negative everywhere?

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Yes! If the leading coefficient is negative (a < 0) and the discriminant is also negative (Δ < 0), then the parabola opens downward and never touches the x-axis, staying negative everywhere.

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