Domain Analysis: Finding Valid Inputs for y = (7/9)x² + 2

Quadratic Functions with Positive Domain Analysis

Find the positive and negative domains of the function:

y=79x2+2 y=\frac{7}{9}x^2+2

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Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Find the positive and negative domains of the function:

y=79x2+2 y=\frac{7}{9}x^2+2

2

Step-by-step solution

To solve this problem, we must determine when the quadratic function y=79x2+2 y = \frac{7}{9}x^2 + 2 is positive or negative.

  • The function is a simple parabola opening upwards because the coefficient of x2 x^2 is positive (79>0 \frac{7}{9} > 0 ). This indicates the graph of the function is positioned above the x-axis or tangent to it if it has roots (but here it clearly has no zero-crossings due to +2 as constant).

  • Clearly, since this expression 79x2 \frac{7}{9}x^2 is always non-negative for any real number x x , plus 2, the value of y y is always positive.

  • For x<0 x < 0 , the term x2 x^2 is still positive, hence y y remains positive. Therefore, there is no negative domain for x<0 x < 0 .

  • For x>0 x > 0 , the quadratic behavior above justifies that y y is always positive across the entire positive domain of x x .

By the analysis above:

x<0:none x < 0 : \text{none}

x>0:all x x > 0 : \text{all } x

Thus, the correct choice is Choice 3:

x<0:none x < 0 : \text{none}

x>0:all x x > 0 : \text{all } x

Therefore, the solution to the problem is x<0: x < 0 : none, x>0: x > 0 : all x x .

3

Final Answer

x<0: x < 0 : none

x>0: x > 0 : all x

Key Points to Remember

Essential concepts to master this topic
  • Parabola Rule: Upward opening parabolas with a>0 a > 0 have minimum values
  • Technique: Find vertex at y=79(0)2+2=2 y = \frac{7}{9}(0)^2 + 2 = 2 minimum
  • Check: Test any x-value: y=79(3)2+2=9>0 y = \frac{7}{9}(3)^2 + 2 = 9 > 0

Common Mistakes

Avoid these frequent errors
  • Confusing domain with range analysis
    Don't analyze where the function crosses zero when asked for positive/negative domains! This gives wrong classifications. The question asks where y-values are positive or negative, not x-values. Always check if y > 0 or y < 0 for the entire function.

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' and 'negative domain' mean?

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These terms ask where the function outputs (y-values) are positive or negative, not the inputs. For y=79x2+2 y = \frac{7}{9}x^2 + 2 , we need to find where y > 0 and where y < 0.

Why is the function always positive?

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Since x20 x^2 \geq 0 for all real numbers, 79x20 \frac{7}{9}x^2 \geq 0 . Adding 2 gives us y2>0 y \geq 2 > 0 . The function never goes below 2!

How do I know the parabola opens upward?

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Look at the coefficient of x2 x^2 ! Since 79>0 \frac{7}{9} > 0 , the parabola opens upward. Negative coefficients make parabolas open downward.

Do I need to find where the function equals zero?

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Not for this problem! We're analyzing where y is positive or negative. Since y=79x2+22 y = \frac{7}{9}x^2 + 2 \geq 2 , the function is always positive and never equals zero.

Why does x < 0 have 'none' for positive domain?

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This is asking where y < 0 when x < 0. Since the function is always positive (y ≥ 2), there are no x-values that make y negative, hence 'none'.

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