Certainly! Let's go through the step-by-step solution to graph the quadratic equation [tex]\( y = x^2 + 6x + 8 \)[/tex]. We will identify the [tex]$y$[/tex]-intercept, the [tex]$x$[/tex]-intercepts, and the vertex of the parabola.
### Step 1: Identify the [tex]$y$[/tex]-intercept
The [tex]$y$[/tex]-intercept of a quadratic equation [tex]\( y = ax^2 + bx + c \)[/tex] is found by setting [tex]\( x = 0 \)[/tex] and solving for [tex]\( y \)[/tex].
[tex]\[
y = (0)^2 + 6(0) + 8 = 8
\][/tex]
So, the [tex]$y$[/tex]-intercept is at [tex]\( (0, 8) \)[/tex].
### Step 2: Identify the [tex]$x$[/tex]-intercepts
The [tex]$x$[/tex]-intercepts of a quadratic equation are the points where [tex]\( y = 0 \)[/tex]. To find these, we solve the equation:
[tex]\[
x^2 + 6x + 8 = 0
\][/tex]
Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1, b = 6, \)[/tex] and [tex]\( c = 8 \)[/tex]:
[tex]\[
x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1}
\][/tex]
[tex]\[
x = \frac{-6 \pm \sqrt{36 - 32}}{2}
\][/tex]
[tex]\[
x = \frac{-6 \pm \sqrt{4}}{2}
\][/tex]
[tex]\[
x = \frac{-6 \pm 2}{2}
\][/tex]
This gives us two solutions:
[tex]\[
x = \frac{-6 + 2}{2} = \frac{-4}{2} = -2
\][/tex]
[tex]\[
x = \frac{-6 - 2}{2} = \frac{-8}{2} = -4
\][/tex]
So, the [tex]$x$[/tex]-intercepts are at [tex]\( (-2, 0) \)[/tex] and [tex]\( (-4, 0) \)[/tex].
### Step 3: Identify the vertex
The vertex of a quadratic equation [tex]\( y = ax^2 + bx + c \)[/tex] occurs at:
[tex]\[
x = -\frac{b}{2a}
\][/tex]
Substituting [tex]\( a = 1 \)[/tex] and [tex]\( b = 6 \)[/tex]:
[tex]\[
x = -\frac{6}{2 \cdot 1} = -3
\][/tex]
Now, calculate the [tex]$y$[/tex]-coordinate by substituting [tex]\( x = -3 \)[/tex] back into the quadratic equation:
[tex]\[
y = (-3)^2 + 6(-3) + 8
\][/tex]
[tex]\[
y = 9 - 18 + 8
\][/tex]
[tex]\[
y = -1
\][/tex]
So, the vertex is at [tex]\( (-3, -1) \)[/tex].
### Step 4: Graphing the equation
1. Plot the [tex]$y$[/tex]-intercept at [tex]\( (0, 8) \)[/tex].
2. Plot the [tex]$x$[/tex]-intercepts at [tex]\( (-2, 0) \)[/tex] and [tex]\( (-4, 0) \)[/tex].
3. Plot the vertex at [tex]\( (-3, -1) \)[/tex].
4. Draw a parabolic curve passing through these points, opening upwards, as the coefficient of [tex]\( x^2 \)[/tex] is positive.
This creates the graph of the quadratic equation [tex]\( y = x^2 + 6x + 8 \)[/tex], showing all the key features:
- [tex]$y$[/tex]-intercept at [tex]\( (0, 8) \)[/tex]
- [tex]$x$[/tex]-intercepts at [tex]\( (-2, 0) \)[/tex] and [tex]\( (-4, 0) \)[/tex]
- Vertex at [tex]\( (-3, -1) \)[/tex]
The graph would look like a parabola opening upwards, with the mentioned points highlighted.
