y = x^2 + 8x + 11

Steps to Solve

1. Identify the quadratic function The given function is a quadratic equation of the form $y = ax^2 + bx + c$.

Here, $a = 1$, $b = 8$, and $c = 11$.

2. Find the vertex of the parabola The vertex can be found using the formula for the x-coordinate of the vertex: $$ x = -\frac{b}{2a} $$ Substituting the values, we get: $$ x = -\frac{8}{2 \cdot 1} = -4 $$

3. Calculate the y-coordinate of the vertex Substitute $x = -4$ back into the original equation to find $y$: $$ y = (-4)^2 + 8(-4) + 11 $$ $$ y = 16 - 32 + 11 = -5 $$

4. Summarize the vertex The vertex of the parabola is at the point $(-4, -5)$.

5. Find the y-intercept The y-intercept occurs when $x = 0$. Substituting $x = 0$ into the equation gives: $$ y = 0^2 + 8(0) + 11 = 11 $$ The y-intercept is $(0, 11)$.

6. Find the x-intercepts (if any) To find the x-intercepts, set $y = 0$: $$ 0 = x^2 + 8x + 11 $$ We can use the quadratic formula: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

Substituting the values: $$ x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot 11}}{2 \cdot 1} $$ $$ x = \frac{-8 \pm \sqrt{64 - 44}}{2} $$ $$ x = \frac{-8 \pm \sqrt{20}}{2} $$ $$ x = \frac{-8 \pm 2\sqrt{5}}{2} $$ $$ x = -4 \pm \sqrt{5} $$

Thus, the x-intercepts are $x = -4 + \sqrt{5}$ and $x = -4 - \sqrt{5}$.