Answer :
Absolutely! Let's start with the given quadratic equation and solve it step-by-step.
The given equation is:
[tex]\[ x^2 + 6x + 8 = 0 \][/tex]
### Step 1: Factoring the Quadratic Equation
To factor this quadratic equation, we need to find two numbers that multiply to the constant term [tex]\(8\)[/tex] and add up to the coefficient of the linear term [tex]\(6\)[/tex].
Let's find such numbers:
- The constant term ([tex]\(c\)[/tex]) is [tex]\(8\)[/tex].
- The coefficient of the linear term ([tex]\(b\)[/tex]) is [tex]\(6\)[/tex].
We need two numbers [tex]\(a\)[/tex] and [tex]\(b\)[/tex] such that:
[tex]\[ a \cdot b = 8 \][/tex]
[tex]\[ a + b = 6 \][/tex]
Testing possible factor pairs of [tex]\(8\)[/tex]:
- [tex]\(1 \cdot 8 = 8\)[/tex] and [tex]\(1 + 8 = 9\)[/tex] (does not work)
- [tex]\(2 \cdot 4 = 8\)[/tex] and [tex]\(2 + 4 = 6\)[/tex] (these are the correct numbers!)
So, the numbers that work are [tex]\(2\)[/tex] and [tex]\(4\)[/tex].
We can now write the quadratic expression as a product of two binomials:
[tex]\[ x^2 + 6x + 8 = (x + 2)(x + 4) = 0 \][/tex]
### Step 2: Solving for [tex]\(x\)[/tex]
Now, set each binomial equal to zero and solve for [tex]\(x\)[/tex]:
[tex]\[ x + 2 = 0 \quad \text{or} \quad x + 4 = 0 \][/tex]
Solving these equations:
[tex]\[ x = -2 \quad \text{and} \quad x = -4 \][/tex]
Thus, the [tex]\(x\)[/tex]-intercepts are:
[tex]\[ x = -2 \quad \text{and} \quad x = -4 \][/tex]
### Step 3: Stating the [tex]\(y\)[/tex]-Intercept
The [tex]\(y\)[/tex]-intercept of a function is the point where the graph intersects the [tex]\(y\)[/tex]-axis. This occurs when [tex]\(x = 0\)[/tex].
To find the [tex]\(y\)[/tex]-intercept, substitute [tex]\(x = 0\)[/tex] into the original equation:
[tex]\[ y = 0^2 + 6(0) + 8 = 8 \][/tex]
So, the [tex]\(y\)[/tex]-intercept is:
[tex]\[ 8 \][/tex]
### Summary
- The [tex]\(x\)[/tex]-intercepts are [tex]\(-4\)[/tex] and [tex]\(-2\)[/tex].
- The [tex]\(y\)[/tex]-intercept is [tex]\(8\)[/tex].
Thus, the detailed solutions are:
- [tex]\( x\)[/tex]-intercepts: [tex]\(-4\)[/tex] and [tex]\(-2\)[/tex]
- [tex]\( y\)[/tex]-intercept: [tex]\(8\)[/tex]
The given equation is:
[tex]\[ x^2 + 6x + 8 = 0 \][/tex]
### Step 1: Factoring the Quadratic Equation
To factor this quadratic equation, we need to find two numbers that multiply to the constant term [tex]\(8\)[/tex] and add up to the coefficient of the linear term [tex]\(6\)[/tex].
Let's find such numbers:
- The constant term ([tex]\(c\)[/tex]) is [tex]\(8\)[/tex].
- The coefficient of the linear term ([tex]\(b\)[/tex]) is [tex]\(6\)[/tex].
We need two numbers [tex]\(a\)[/tex] and [tex]\(b\)[/tex] such that:
[tex]\[ a \cdot b = 8 \][/tex]
[tex]\[ a + b = 6 \][/tex]
Testing possible factor pairs of [tex]\(8\)[/tex]:
- [tex]\(1 \cdot 8 = 8\)[/tex] and [tex]\(1 + 8 = 9\)[/tex] (does not work)
- [tex]\(2 \cdot 4 = 8\)[/tex] and [tex]\(2 + 4 = 6\)[/tex] (these are the correct numbers!)
So, the numbers that work are [tex]\(2\)[/tex] and [tex]\(4\)[/tex].
We can now write the quadratic expression as a product of two binomials:
[tex]\[ x^2 + 6x + 8 = (x + 2)(x + 4) = 0 \][/tex]
### Step 2: Solving for [tex]\(x\)[/tex]
Now, set each binomial equal to zero and solve for [tex]\(x\)[/tex]:
[tex]\[ x + 2 = 0 \quad \text{or} \quad x + 4 = 0 \][/tex]
Solving these equations:
[tex]\[ x = -2 \quad \text{and} \quad x = -4 \][/tex]
Thus, the [tex]\(x\)[/tex]-intercepts are:
[tex]\[ x = -2 \quad \text{and} \quad x = -4 \][/tex]
### Step 3: Stating the [tex]\(y\)[/tex]-Intercept
The [tex]\(y\)[/tex]-intercept of a function is the point where the graph intersects the [tex]\(y\)[/tex]-axis. This occurs when [tex]\(x = 0\)[/tex].
To find the [tex]\(y\)[/tex]-intercept, substitute [tex]\(x = 0\)[/tex] into the original equation:
[tex]\[ y = 0^2 + 6(0) + 8 = 8 \][/tex]
So, the [tex]\(y\)[/tex]-intercept is:
[tex]\[ 8 \][/tex]
### Summary
- The [tex]\(x\)[/tex]-intercepts are [tex]\(-4\)[/tex] and [tex]\(-2\)[/tex].
- The [tex]\(y\)[/tex]-intercept is [tex]\(8\)[/tex].
Thus, the detailed solutions are:
- [tex]\( x\)[/tex]-intercepts: [tex]\(-4\)[/tex] and [tex]\(-2\)[/tex]
- [tex]\( y\)[/tex]-intercept: [tex]\(8\)[/tex]
