Westonci.ca offers fast, accurate answers to your questions. Join our community and get the insights you need now. Experience the ease of finding quick and accurate answers to your questions from professionals on our platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
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.
### 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.
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.
