To determine which parabola will have exactly one real solution when intersected by the line [tex]\( y = x - 5 \)[/tex], we need to determine the points of intersection between the given parabolas and the line.
For a parabola to have one real solution (i.e., touch the line at exactly one point), the quadratic equation formed by setting the two equations equal must have a discriminant of zero.
1. Equation and Parabola:
Given parabola: [tex]\( y = x^2 + x - 4 \)[/tex]
Given line: [tex]\( y = x - 5 \)[/tex]
2. Substitute [tex]\( y = x - 5 \)[/tex] into the Parabola Equation:
Set the equations equal to each other to find the points of intersection:
[tex]\[
x^2 + x - 4 = x - 5
\][/tex]
3. Simplify the Equation:
Rearrange terms to form a standard quadratic equation:
[tex]\[
x^2 + x - 4 - (x - 5) = 0
\][/tex]
[tex]\[
x^2 + x - 4 - x + 5 = 0
\][/tex]
[tex]\[
x^2 + 1 = 0
\][/tex]
4. Solve the Quadratic Equation:
The simplified quadratic equation is:
[tex]\[
x^2 + 1 = 0
\][/tex]
To have one real solution, the discriminant [tex]\((b^2 - 4ac)\)[/tex] should be equal to zero.
5. Find the Discriminant:
For the equation [tex]\( x^2 + 1 = 0 \)[/tex]:
[tex]\( a = 1 \)[/tex], [tex]\( b = 0 \)[/tex], [tex]\( c = 1 \)[/tex]
The discriminant is given by:
[tex]\[
\Delta = b^2 - 4ac = 0^2 - 4 \cdot 1 \cdot 1 = 0 - 4 = -4
\][/tex]
Since the discriminant [tex]\(\Delta\)[/tex] is negative (i.e., [tex]\(-4\)[/tex]), this means there are no real solutions to the quadratic equation [tex]\( x^2 + 1 = 0 \)[/tex].
Therefore, the correct answer is that the parabola:
[tex]\[
y = x^2 + x - 4
\][/tex]
will touch the line [tex]\( y = x - 5 \)[/tex] at exactly one point.
