Answer :
To effectively graph the parabola given by the equation [tex]\( y = 2x^2 + 8x - 12 \)[/tex], it's helpful to identify several key features of the parabola: the vertex, axis of symmetry, [tex]\(x\)[/tex]-intercepts, and [tex]\(y\)[/tex]-intercept.
### 1. Finding the Vertex
The vertex of a parabola given by [tex]\( y = ax^2 + bx + c \)[/tex] can be found using the formula for the [tex]\(x\)[/tex]-coordinate of the vertex:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Here, [tex]\( a = 2 \)[/tex], [tex]\( b = 8 \)[/tex], and [tex]\( c = -12 \)[/tex]. Plugging in these values:
[tex]\[ x = -\frac{8}{2 \times 2} = -\frac{8}{4} = -2 \][/tex]
Now, substitute [tex]\( x = -2 \)[/tex] into the original equation to find the [tex]\( y \)[/tex]-coordinate:
[tex]\[ y = 2(-2)^2 + 8(-2) - 12 \][/tex]
[tex]\[ y = 2(4) + (-16) - 12 \][/tex]
[tex]\[ y = 8 - 16 - 12 \][/tex]
[tex]\[ y = -20 \][/tex]
So, the vertex of the parabola is [tex]\( (-2, -20) \)[/tex].
### 2. Axis of Symmetry
The axis of symmetry is the vertical line that passes through the vertex. Therefore, it is given by the equation:
[tex]\[ x = -2 \][/tex]
### 3. Finding the [tex]\( y \)[/tex]-Intercept
The [tex]\( y \)[/tex]-intercept is found by setting [tex]\( x = 0 \)[/tex] in the original equation and solving for [tex]\( y \)[/tex]:
[tex]\[ y = 2(0)^2 + 8(0) - 12 \][/tex]
[tex]\[ y = -12 \][/tex]
So, the [tex]\( y \)[/tex]-intercept is [tex]\( (0, -12) \)[/tex].
### 4. Finding the [tex]\( x \)[/tex]-Intercepts
The [tex]\( x \)[/tex]-intercepts are found by setting [tex]\( y = 0 \)[/tex] and solving the resulting quadratic equation:
[tex]\[ 2x^2 + 8x - 12 = 0 \][/tex]
We use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ a = 2, \; b = 8, \; c = -12 \][/tex]
[tex]\[ \text{Discriminant} = b^2 - 4ac = 8^2 - 4(2)(-12) \][/tex]
[tex]\[ \text{Discriminant} = 64 + 96 \][/tex]
[tex]\[ \text{Discriminant} = 160 \][/tex]
Since the discriminant is positive, we have two real roots:
[tex]\[ x = \frac{-8 \pm \sqrt{160}}{4} \][/tex]
[tex]\[ x = \frac{-8 \pm 4\sqrt{10}}{4} \][/tex]
[tex]\[ x = -2 \pm \sqrt{10} \][/tex]
This gives the [tex]\( x \)[/tex]-intercepts:
[tex]\[ x = -2 + \sqrt{10} \approx 1.162 \][/tex]
[tex]\[ x = -2 - \sqrt{10} \approx -5.162 \][/tex]
So, the [tex]\( x \)[/tex]-intercepts are approximately:
[tex]\[ \left(1.162, 0\right) \; \text{and} \; \left(-5.162, 0\right) \][/tex]
### Summary
Using the above information, we have:
- Vertex: [tex]\((-2, -20)\)[/tex]
- Axis of Symmetry: [tex]\( x = -2 \)[/tex]
- [tex]\( y \)[/tex]-Intercept: [tex]\((0, -12)\)[/tex]
- [tex]\( x \)[/tex]-Intercepts: [tex]\(\left(1.162, 0\right) \; \text{and} \; \left(-5.162, 0\right)\)[/tex]
These details will allow you to graph the parabola accurately.
In conclusion, the vertex of the parabola [tex]\( y = 2x^2 + 8x - 12 \)[/tex] is:
[tex]\[ (-2, -20) \][/tex]
### 1. Finding the Vertex
The vertex of a parabola given by [tex]\( y = ax^2 + bx + c \)[/tex] can be found using the formula for the [tex]\(x\)[/tex]-coordinate of the vertex:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Here, [tex]\( a = 2 \)[/tex], [tex]\( b = 8 \)[/tex], and [tex]\( c = -12 \)[/tex]. Plugging in these values:
[tex]\[ x = -\frac{8}{2 \times 2} = -\frac{8}{4} = -2 \][/tex]
Now, substitute [tex]\( x = -2 \)[/tex] into the original equation to find the [tex]\( y \)[/tex]-coordinate:
[tex]\[ y = 2(-2)^2 + 8(-2) - 12 \][/tex]
[tex]\[ y = 2(4) + (-16) - 12 \][/tex]
[tex]\[ y = 8 - 16 - 12 \][/tex]
[tex]\[ y = -20 \][/tex]
So, the vertex of the parabola is [tex]\( (-2, -20) \)[/tex].
### 2. Axis of Symmetry
The axis of symmetry is the vertical line that passes through the vertex. Therefore, it is given by the equation:
[tex]\[ x = -2 \][/tex]
### 3. Finding the [tex]\( y \)[/tex]-Intercept
The [tex]\( y \)[/tex]-intercept is found by setting [tex]\( x = 0 \)[/tex] in the original equation and solving for [tex]\( y \)[/tex]:
[tex]\[ y = 2(0)^2 + 8(0) - 12 \][/tex]
[tex]\[ y = -12 \][/tex]
So, the [tex]\( y \)[/tex]-intercept is [tex]\( (0, -12) \)[/tex].
### 4. Finding the [tex]\( x \)[/tex]-Intercepts
The [tex]\( x \)[/tex]-intercepts are found by setting [tex]\( y = 0 \)[/tex] and solving the resulting quadratic equation:
[tex]\[ 2x^2 + 8x - 12 = 0 \][/tex]
We use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ a = 2, \; b = 8, \; c = -12 \][/tex]
[tex]\[ \text{Discriminant} = b^2 - 4ac = 8^2 - 4(2)(-12) \][/tex]
[tex]\[ \text{Discriminant} = 64 + 96 \][/tex]
[tex]\[ \text{Discriminant} = 160 \][/tex]
Since the discriminant is positive, we have two real roots:
[tex]\[ x = \frac{-8 \pm \sqrt{160}}{4} \][/tex]
[tex]\[ x = \frac{-8 \pm 4\sqrt{10}}{4} \][/tex]
[tex]\[ x = -2 \pm \sqrt{10} \][/tex]
This gives the [tex]\( x \)[/tex]-intercepts:
[tex]\[ x = -2 + \sqrt{10} \approx 1.162 \][/tex]
[tex]\[ x = -2 - \sqrt{10} \approx -5.162 \][/tex]
So, the [tex]\( x \)[/tex]-intercepts are approximately:
[tex]\[ \left(1.162, 0\right) \; \text{and} \; \left(-5.162, 0\right) \][/tex]
### Summary
Using the above information, we have:
- Vertex: [tex]\((-2, -20)\)[/tex]
- Axis of Symmetry: [tex]\( x = -2 \)[/tex]
- [tex]\( y \)[/tex]-Intercept: [tex]\((0, -12)\)[/tex]
- [tex]\( x \)[/tex]-Intercepts: [tex]\(\left(1.162, 0\right) \; \text{and} \; \left(-5.162, 0\right)\)[/tex]
These details will allow you to graph the parabola accurately.
In conclusion, the vertex of the parabola [tex]\( y = 2x^2 + 8x - 12 \)[/tex] is:
[tex]\[ (-2, -20) \][/tex]