Solve y=x²+2x+2: Finding Where Function Values Are Positive

Quadratic Functions with Vertex Analysis

Look at the function below:

y=x2+2x+2 y=x^2+2x+2

Determine for which values of x x the following is true:

f(x)>0 f\left(x\right)>0

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Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Look at the function below:

y=x2+2x+2 y=x^2+2x+2

Determine for which values of x x the following is true:

f(x)>0 f\left(x\right)>0

2

Step-by-step solution

To find for which values of x x the function y=x2+2x+2 y = x^2 + 2x + 2 is positive, we'll begin by analyzing the quadratic expression.

First, let's complete the square for the quadratic function:

y=x2+2x+2 y = x^2 + 2x + 2

To complete the square, take the coefficient of the x x term (which is 2), divide it by 2 to get 1, and then square it to obtain 1. Add and subtract this value inside the expression:

y=(x2+2x+1)+21 y = (x^2 + 2x + 1) + 2 - 1

This simplifies to:

y=(x+1)2+1 y = (x + 1)^2 + 1

Now, this function y=(x+1)2+1 y = (x+1)^2 + 1 shows a parabola in the vertex form (xh)2+k(x-h)^2 + k, where the vertex is at (1,1)(-1, 1) and opens upwards, as the coefficient of the squared term is positive.

This indicates that the minimum point, (vertex) \text{(vertex)} , is at y=1 y = 1 , which is above the x-axis. As such, the function y=(x+1)2+1 y = (x+1)^2 + 1 will be positive for all x x since the entire curve lies above the x-axis and the value of y y is always greater than zero.

Therefore, the function y=x2+2x+2 y = x^2 + 2x + 2 is positive for all real values of x x .

Hence, the solution is that the function is positive for all values of x x .

3

Final Answer

The function is positive for all values of x x .

Key Points to Remember

Essential concepts to master this topic
  • Vertex Form: Complete the square to identify minimum value
  • Technique: (x+1)2+1 (x+1)^2 + 1 shows vertex at (-1, 1)
  • Check: Minimum value is 1, so function is always positive ✓

Common Mistakes

Avoid these frequent errors
  • Trying to find x-intercepts by setting y = 0
    Don't solve x2+2x+2=0 x^2 + 2x + 2 = 0 to find where it's positive = no real solutions exist! This leads to confusion about the function's behavior. Always complete the square to find the vertex and determine the minimum value.

Practice Quiz

Test your knowledge with interactive questions

The graph of the function below does not intersect the \( x \)-axis.

The parabola's vertex is marked A.

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

AAAX

FAQ

Everything you need to know about this question

Why doesn't this quadratic have x-intercepts?

+

The discriminant b24ac=48=4 b^2 - 4ac = 4 - 8 = -4 is negative, so there are no real x-intercepts. The parabola stays completely above the x-axis!

How do I know the parabola opens upward?

+

Look at the coefficient of x2 x^2 . Since it's positive (+1), the parabola opens upward like a smile, not downward.

What does completing the square actually show me?

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Completing the square gives you vertex form: y=(x+1)2+1 y = (x+1)^2 + 1 . This tells you the lowest point is at (-1, 1), so the function never goes below y = 1.

Can a quadratic function be positive for all x values?

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Yes! When a parabola opens upward and its vertex is above the x-axis, the entire function stays positive. This happens when the minimum value is greater than zero.

How do I verify that f(x) > 0 for all x?

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Since y=(x+1)2+1 y = (x+1)^2 + 1 and (x+1)20 (x+1)^2 ≥ 0 for all real x, we have y1>0 y ≥ 1 > 0 . The function is always at least 1!

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