To find the points of intersection for the given system of equations:
[tex]\[ y = 2x^2 \][/tex]
[tex]\[ y = -3x - 1 \][/tex]
we need to set the two equations equal to each other because we are searching for the [tex]\(x\)[/tex] values where the [tex]\(y\)[/tex] values from both equations are the same.
Step 1: Set the equations equal to each other:
[tex]\[ 2x^2 = -3x - 1 \][/tex]
Step 2: Rearrange the equation to standard quadratic form:
[tex]\[ 2x^2 + 3x + 1 = 0 \][/tex]
Step 3: Solve the quadratic equation [tex]\(2x^2 + 3x + 1 = 0\)[/tex]. We can solve using the quadratic formula, [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 2\)[/tex], [tex]\(b = 3\)[/tex], and [tex]\(c = 1\)[/tex].
First, calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = 3^2 - 4 \cdot 2 \cdot 1 \][/tex]
[tex]\[ \Delta = 9 - 8 \][/tex]
[tex]\[ \Delta = 1 \][/tex]
Step 4: Using the quadratic formula:
[tex]\[ x = \frac{-3 \pm \sqrt{1}}{2 \cdot 2} \][/tex]
[tex]\[ x = \frac{-3 \pm 1}{4} \][/tex]
This gives us two solutions for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{-3 + 1}{4} = \frac{-2}{4} = -\frac{1}{2} \][/tex]
[tex]\[ x = \frac{-3 - 1}{4} = \frac{-4}{4} = -1 \][/tex]
So the solutions for [tex]\(x\)[/tex] are [tex]\(x = -1\)[/tex] and [tex]\(x = -\frac{1}{2}\)[/tex].
Step 5: Find the corresponding [tex]\(y\)[/tex] values by substituting each [tex]\(x\)[/tex] value back into either of the original equations. We'll use [tex]\(y = 2x^2\)[/tex] for simplicity:
For [tex]\(x = -1\)[/tex]:
[tex]\[ y = 2(-1)^2 = 2 \cdot 1 = 2 \][/tex]
For [tex]\(x = -\frac{1}{2}\)[/tex]:
[tex]\[ y = 2\left(-\frac{1}{2}\right)^2 = 2 \cdot \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \][/tex]
Thus, the points of intersection are:
[tex]\[ (-1, 2) \][/tex]
[tex]\[ \left(-\frac{1}{2}, \frac{1}{2}\right) \][/tex]
In conclusion, the system of equations intersects at the points [tex]\((-1, 2)\)[/tex] and [tex]\(\left(-\frac{1}{2}, \frac{1}{2}\right)\)[/tex].
