To solve the system of equations:
- \( y = 14(x - 1)^2 - 3 \)
- \( y = -12x - 12 \)
we can set them equal to each other to find the intersection points:
\[ 14(x - 1)^2 - 3 = -12x - 12 \]
Now, we will simplify and rearrange:
\[ 14(x - 1)^2 + 12x + 9 = 0 \]
Next, we expand \(14(x - 1)^2\):
\[ 14(x^2 - 2x + 1) = 14x^2 - 28x + 14 \]
So we have:
\[ 14x^2 - 28x + 14 + 12x + 9 = 0 \]
Combining like terms gives:
\[ 14x^2 - 16x + 23 = 0 \]
Now we will apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Here, \( a = 14 \), \( b = -16 \), and \( c = 23 \).
First, we calculate the discriminant:
\[ b^2 - 4ac = (-16)^2 - 4 \cdot 14 \cdot 23 = 256 - 1288 = -1032 \]
Since the discriminant is negative, this means that there are no real solutions for this system of equations.
However, the problem states that there are solutions present on the given coordinate grid. This may imply that there was a visual aspect to the solution process. Based on the context, if we assume that the tuitional value of x involves some limited range, a decision about when to solve computationally or graphically spirals in might've produced different outcomes.
If we are indeed given the solutions, please provide any of them, and I can assist in framing them accordingly. Otherwise, working with the equations directly does imply that the intersection is nonexistent in real-number fields.
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