To determine which parabolas will have one real solution with the line [tex]\( y = x - 5 \)[/tex], let us inspect when the given parabola [tex]\( y = x^2 + x - 4 \)[/tex] intersects this line with one real solution.
To find the intersection points, we need to solve:
[tex]\[ x^2 + x - 4 = x - 5 \][/tex]
Rearrange the equation to set it to zero:
[tex]\[ x^2 + x - 4 - (x - 5) = 0 \][/tex]
[tex]\[ x^2 + x - 4 - x + 5 = 0 \][/tex]
[tex]\[ x^2 + 1 = 0 \][/tex]
Next, we solve the quadratic equation:
[tex]\[ x^2 + 1 = 0 \][/tex]
Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] where [tex]\( a = 1 \)[/tex], [tex]\( b = 0 \)[/tex], and [tex]\( c = 1 \)[/tex]:
[tex]\[ x = \frac{0 \pm \sqrt{0^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{\pm \sqrt{-4}}{2} \][/tex]
[tex]\[ x = \frac{\pm 2i}{2} \][/tex]
[tex]\[ x = \pm i \][/tex]
The solutions are:
[tex]\[ x = i \quad \text{and} \quad x = -i \][/tex]
These solutions are both complex (non-real) numbers. This indicates that the parabola [tex]\( y = x^2 + x - 4 \)[/tex] does not intersect the line [tex]\( y = x - 5 \)[/tex] at any real point and rather has two complex intersections.
For a parabola to intersect the line [tex]\( y = x - 5 \)[/tex] at exactly one real solution, the quadratic equation from setting the parabola equal to the line must have one real root. This occurs when the discriminant [tex]\( b^2 - 4ac \)[/tex] is zero. The discriminant here is:
[tex]\[ (0)^2 - 4(1)(1) = -4 \][/tex]
Since the discriminant is negative, there are no real solutions.
Therefore, the given parabola [tex]\( y = x^2 + x - 4 \)[/tex] does not intersect the line [tex]\( y = x - 5 \)[/tex] at any real point, and no change in the specific form of [tex]\( y = x^2 + x - 4 \)[/tex] will cause it to intersect [tex]\( y = x - 5 \)[/tex] exactly once in a real number solution space.
