1. **State the problem:** We want to prove that $$2x^2 + 2x > -4$$ for all real values of $$x$$, and then explain why the graph of $$y = 2x^2 + 2x + 4$$ has no x-intercepts. 2. **Rewrite the inequality:** Start with the given inequality: $$2x^2 + 2x > -4$$ Divide both sides by 2: $$\cancel{2}x^2 + \cancel{2}x > \frac{-4}{\cancel{2}}$$ which simplifies to $$x^2 + x > -2$$ 3. **Complete the square:** Recall the formula for completing the square: $$x^2 + bx = (x + \frac{b}{2})^2 - \left(\frac{b}{2}\right)^2$$ Here, $$b = 1$$, so $$x^2 + x = \left(x + \frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 = \left(x + \frac{1}{2}\right)^2 - \frac{1}{4}$$ 4. **Substitute back into the inequality:** $$\left(x + \frac{1}{2}\right)^2 - \frac{1}{4} > -2$$ Add $$\frac{1}{4}$$ to both sides: $$\left(x + \frac{1}{2}\right)^2 > -2 + \frac{1}{4} = -\frac{7}{4}$$ 5. **Analyze the inequality:** Since $$\left(x + \frac{1}{2}\right)^2$$ is a square, it is always $$\geq 0$$ for all real $$x$$. Because $$0 > -\frac{7}{4}$$, the inequality $$\left(x + \frac{1}{2}\right)^2 > -\frac{7}{4}$$ is true for all real $$x$$. 6. **Multiply back by 2:** Recall the original inequality was $$2x^2 + 2x > -4$$ which we have shown is true for all real $$x$$. 7. **Explain why $$y = 2x^2 + 2x + 4$$ has no x-intercepts:** The x-intercepts occur where $$y=0$$: $$2x^2 + 2x + 4 = 0$$ Rewrite as $$2x^2 + 2x = -4$$ But from step 6, we know $$2x^2 + 2x > -4$$ for all $$x$$, so the left side is always greater than $$-4$$. Therefore, the equation $$2x^2 + 2x + 4 = 0$$ has no real solutions, meaning the graph does not cross the x-axis and has no x-intercepts. **Final answer:** $$2x^2 + 2x > -4$$ for all real $$x$$, so $$y = 2x^2 + 2x + 4$$ has no x-intercepts.