y = x² + 8x + 11

Steps to Solve

1. Identify the Vertex of the Parabola

The given equation is in the standard quadratic form $y = ax^2 + bx + c$. Here, $a = 1$, $b = 8$, and $c = 11$. To find the vertex, we use the formula for the $x$-coordinate of the vertex:

$$ x_v = -\frac{b}{2a} $$

Substituting the values:

$$ x_v = -\frac{8}{2 \cdot 1} = -4 $$

2. Calculate the $y$-coordinate of the Vertex

Now we plug $x_v = -4$ back into the original equation to find the $y$-coordinate:

$$ y_v = (-4)^2 + 8(-4) + 11 $$

Calculating this step by step:

• $(-4)^2 = 16$
• $8 \cdot (-4) = -32$
• Therefore,

$$ y_v = 16 - 32 + 11 = -5 $$

So, the vertex is at the point $(-4, -5)$.

3. Determine the Axis of Symmetry

The axis of symmetry is a vertical line that passes through the vertex. Its equation is:

$$ x = -4 $$

4. Find the $y$-intercept

To find the $y$-intercept, set $x = 0$ in the original equation:

$$ y = 0^2 + 8(0) + 11 = 11 $$

So, the $y$-intercept is the point $(0, 11)$.

5. Find the $x$-intercepts

To find the $x$-intercepts, set $y = 0$:

$$ 0 = x^2 + 8x + 11 $$

Using the quadratic formula:

$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

Substituting values:

$$ x = \frac{-8 \pm \sqrt{8^2 - 4(1)(11)}}{2(1)} = \frac{-8 \pm \sqrt{64 - 44}}{2} = \frac{-8 \pm \sqrt{20}}{2} $$ $$ \sqrt{20} = 2\sqrt{5} $$ Thus,

$$ x = \frac{-8 \pm 2\sqrt{5}}{2} = -4 \pm \sqrt{5} $$

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