To solve the system of equations given by:
[tex]\[
\begin{array}{l}
y=3x \\
y=x^2-4
\end{array}
\][/tex]
we need to find the points of intersection between the two curves. Here's the step-by-step method to do that:
1. Set the Equations Equal:
Since both expressions for [tex]\( y \)[/tex] are equal to each other, we set them equal to find the [tex]\( x \)[/tex]-values where the curves intersect:
[tex]\[
3x = x^2 - 4
\][/tex]
2. Rearrange into a Standard Quadratic Equation:
Move all terms to one side of the equation to set it equal to zero:
[tex]\[
x^2 - 3x - 4 = 0
\][/tex]
3. Solve the Quadratic Equation:
To solve [tex]\( x^2 - 3x - 4 = 0 \)[/tex], we can use the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = -3 \)[/tex], and [tex]\( c = -4 \)[/tex].
4. Calculate the Discriminant:
The discriminant is calculated as:
[tex]\[
\Delta = b^2 - 4ac = (-3)^2 - 4(1)(-4) = 9 + 16 = 25
\][/tex]
5. Find the Roots:
Using the quadratic formula:
[tex]\[
x = \frac{3 \pm \sqrt{25}}{2(1)} = \frac{3 \pm 5}{2}
\][/tex]
This gives us two solutions:
[tex]\[
x = \frac{3 + 5}{2} = \frac{8}{2} = 4
\][/tex]
[tex]\[
x = \frac{3 - 5}{2} = \frac{-2}{2} = -1
\][/tex]
6. Find the Corresponding [tex]\( y \)[/tex]-Values:
Substitute [tex]\( x = 4 \)[/tex] and [tex]\( x = -1 \)[/tex] into the equation [tex]\( y = 3x \)[/tex] to find the corresponding [tex]\( y \)[/tex]-values:
[tex]\[
\text{For } x = 4: \quad y = 3(4) = 12
\][/tex]
[tex]\[
\text{For } x = -1: \quad y = 3(-1) = -3
\][/tex]
Therefore, the points of intersection are [tex]\( (4, 12) \)[/tex] and [tex]\( (-1, -3) \)[/tex].
By comparing this with the provided options, the correct answer is:
C. [tex]\( (-1,-3) \)[/tex] and [tex]\( (4,12) \)[/tex].
