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Certainly! Let's delve into the function [tex]\( y = 2x^2 + 2x - 12 \)[/tex] and analyze it step-by-step in order to understand its properties and behavior.
### Step 1: Factoring and Finding Roots
To understand where the function [tex]\( y \)[/tex] crosses the x-axis, we can find the roots of the quadratic equation [tex]\( 2x^2 + 2x - 12 = 0 \)[/tex].
[tex]\[ y = 2x^2 + 2x - 12 \][/tex]
We can use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] where [tex]\( a = 2 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -12 \)[/tex].
1. [tex]\[ b^2 - 4ac = 2^2 - 4(2)(-12) = 4 + 96 = 100 \][/tex]
2. [tex]\[ \sqrt{b^2 - 4ac} = \sqrt{100} = 10 \][/tex]
3. [tex]\[ x = \frac{-2 \pm 10}{4} \][/tex]
Thus, we have two solutions:
[tex]\[ x_1 = \frac{-2 + 10}{4} = 2 \][/tex]
[tex]\[ x_2 = \frac{-2 - 10}{4} = -3 \][/tex]
So, the roots of the equation [tex]\( 2x^2 + 2x - 12 = 0 \)[/tex] are [tex]\( x = 2 \)[/tex] and [tex]\( x = -3 \)[/tex].
### Step 2: Analyzing the Vertex
The vertex of a parabola [tex]\( y = ax^2 + bx + c \)[/tex] can be found using the formula [tex]\( x = -\frac{b}{2a} \)[/tex].
Here, [tex]\( a = 2 \)[/tex] and [tex]\( b = 2 \)[/tex]:
[tex]\[ x = -\frac{2}{2(2)} = -\frac{1}{2} \][/tex]
To find the y-coordinate of the vertex, we substitute [tex]\( x = -\frac{1}{2} \)[/tex] back into the original equation:
[tex]\[ y = 2 \left(-\frac{1}{2}\right)^2 + 2 \left(-\frac{1}{2}\right) - 12 \][/tex]
[tex]\[ y = 2 \left(\frac{1}{4}\right) - 1 - 12 \][/tex]
[tex]\[ y = \frac{1}{2} - 1 - 12 \][/tex]
[tex]\[ y = - \frac{1}{2} - 12 \][/tex]
[tex]\[ y = -12.5 \][/tex]
So, the vertex of the parabola is at [tex]\( \left( -\frac{1}{2}, -12.5 \right) \)[/tex].
### Step 3: Graphing the Function
To graph the parabola, we consider the roots, the vertex, and the general shape of the parabola.
- The parabola opens upwards since the coefficient of [tex]\( x^2 \)[/tex] is positive (i.e., [tex]\( a > 0 \)[/tex]).
- The roots or x-intercepts are at [tex]\( x = 2 \)[/tex] and [tex]\( x = -3 \)[/tex].
- The vertex is at [tex]\( \left( -\frac{1}{2}, -12.5 \right) \)[/tex].
### Summary
The function [tex]\( y = 2x^2 + 2x - 12 \)[/tex] has:
- Roots at [tex]\( x = 2 \)[/tex] and [tex]\( x = -3 \)[/tex]
- A vertex at [tex]\( \left( -\frac{1}{2}, -12.5 \right) \)[/tex]
- Parabola opens upwards.
These characteristics tell us the essential features of the function [tex]\( y \)[/tex].
### Step 1: Factoring and Finding Roots
To understand where the function [tex]\( y \)[/tex] crosses the x-axis, we can find the roots of the quadratic equation [tex]\( 2x^2 + 2x - 12 = 0 \)[/tex].
[tex]\[ y = 2x^2 + 2x - 12 \][/tex]
We can use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] where [tex]\( a = 2 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -12 \)[/tex].
1. [tex]\[ b^2 - 4ac = 2^2 - 4(2)(-12) = 4 + 96 = 100 \][/tex]
2. [tex]\[ \sqrt{b^2 - 4ac} = \sqrt{100} = 10 \][/tex]
3. [tex]\[ x = \frac{-2 \pm 10}{4} \][/tex]
Thus, we have two solutions:
[tex]\[ x_1 = \frac{-2 + 10}{4} = 2 \][/tex]
[tex]\[ x_2 = \frac{-2 - 10}{4} = -3 \][/tex]
So, the roots of the equation [tex]\( 2x^2 + 2x - 12 = 0 \)[/tex] are [tex]\( x = 2 \)[/tex] and [tex]\( x = -3 \)[/tex].
### Step 2: Analyzing the Vertex
The vertex of a parabola [tex]\( y = ax^2 + bx + c \)[/tex] can be found using the formula [tex]\( x = -\frac{b}{2a} \)[/tex].
Here, [tex]\( a = 2 \)[/tex] and [tex]\( b = 2 \)[/tex]:
[tex]\[ x = -\frac{2}{2(2)} = -\frac{1}{2} \][/tex]
To find the y-coordinate of the vertex, we substitute [tex]\( x = -\frac{1}{2} \)[/tex] back into the original equation:
[tex]\[ y = 2 \left(-\frac{1}{2}\right)^2 + 2 \left(-\frac{1}{2}\right) - 12 \][/tex]
[tex]\[ y = 2 \left(\frac{1}{4}\right) - 1 - 12 \][/tex]
[tex]\[ y = \frac{1}{2} - 1 - 12 \][/tex]
[tex]\[ y = - \frac{1}{2} - 12 \][/tex]
[tex]\[ y = -12.5 \][/tex]
So, the vertex of the parabola is at [tex]\( \left( -\frac{1}{2}, -12.5 \right) \)[/tex].
### Step 3: Graphing the Function
To graph the parabola, we consider the roots, the vertex, and the general shape of the parabola.
- The parabola opens upwards since the coefficient of [tex]\( x^2 \)[/tex] is positive (i.e., [tex]\( a > 0 \)[/tex]).
- The roots or x-intercepts are at [tex]\( x = 2 \)[/tex] and [tex]\( x = -3 \)[/tex].
- The vertex is at [tex]\( \left( -\frac{1}{2}, -12.5 \right) \)[/tex].
### Summary
The function [tex]\( y = 2x^2 + 2x - 12 \)[/tex] has:
- Roots at [tex]\( x = 2 \)[/tex] and [tex]\( x = -3 \)[/tex]
- A vertex at [tex]\( \left( -\frac{1}{2}, -12.5 \right) \)[/tex]
- Parabola opens upwards.
These characteristics tell us the essential features of the function [tex]\( y \)[/tex].
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