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To find which parabola will have one real solution with the line [tex]\( y = x - 5 \)[/tex], we need to determine under what conditions the line and the parabola intersect exactly once.
Here are the detailed steps to find the parabola:
1. Define the Equations:
- The equation of the line is given as [tex]\( y = x - 5 \)[/tex].
- The general form of the parabola we're considering is [tex]\( y = ax^2 + bx + c \)[/tex], where [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] are constants.
2. Set the Equations Equal:
- To find the intersection points, set the equations of the line and the parabola equal to each other:
[tex]\[ x - 5 = ax^2 + bx + c \][/tex]
- Rearrange this to form a standard quadratic equation:
[tex]\[ ax^2 + (b - 1)x + (c + 5) = 0 \][/tex]
3. Condition for One Real Solution:
- A quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] has exactly one real solution (a repeated root) if and only if its discriminant is zero. The discriminant [tex]\(\Delta\)[/tex] of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by [tex]\( b^2 - 4ac \)[/tex].
- For our quadratic equation [tex]\( ax^2 + (b - 1)x + (c + 5) = 0 \)[/tex], the discriminant becomes:
[tex]\[ (b-1)^2 - 4a(c + 5) \][/tex]
4. Set the Discriminant to Zero:
- To have one real solution, set the discriminant equal to zero:
[tex]\[ (b-1)^2 - 4a(c + 5) = 0 \][/tex]
5. Solve for [tex]\( c \)[/tex]:
- Solving the equation for [tex]\( c \)[/tex], we get:
[tex]\[ c = -5 + \frac{(b - 1)^2}{4a} \][/tex]
6. Specific Example:
- Let's find [tex]\( c \)[/tex] for specific values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]. If we set [tex]\( a = 1 \)[/tex] and [tex]\( b = 0 \)[/tex], we can substitute these values into our solution for [tex]\( c \)[/tex]:
[tex]\[ c = -5 + \frac{(0 - 1)^2}{4 \times 1} = -5 + \frac{1}{4} = -5 + 0.25 = -4.75 \][/tex]
- So, for [tex]\( a = 1 \)[/tex] and [tex]\( b = 0 \)[/tex], we have [tex]\( c = -4.75 \)[/tex].
7. Conclusion:
- The parabola given by the equation [tex]\( y = x^2 - 4.75 \)[/tex] will intersect the line [tex]\( y = x - 5 \)[/tex] exactly once, providing us with one real solution.
By following these steps, we can see that the conditions under which a parabola will have exactly one real intersection point with the line [tex]\( y = x - 5 \)[/tex] are given by [tex]\( c = -5 + \frac{(b-1)^2}{4a} \)[/tex], and an example parabola that meets this condition is [tex]\( y = x^2 - 4.75 \)[/tex].
Here are the detailed steps to find the parabola:
1. Define the Equations:
- The equation of the line is given as [tex]\( y = x - 5 \)[/tex].
- The general form of the parabola we're considering is [tex]\( y = ax^2 + bx + c \)[/tex], where [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] are constants.
2. Set the Equations Equal:
- To find the intersection points, set the equations of the line and the parabola equal to each other:
[tex]\[ x - 5 = ax^2 + bx + c \][/tex]
- Rearrange this to form a standard quadratic equation:
[tex]\[ ax^2 + (b - 1)x + (c + 5) = 0 \][/tex]
3. Condition for One Real Solution:
- A quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] has exactly one real solution (a repeated root) if and only if its discriminant is zero. The discriminant [tex]\(\Delta\)[/tex] of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by [tex]\( b^2 - 4ac \)[/tex].
- For our quadratic equation [tex]\( ax^2 + (b - 1)x + (c + 5) = 0 \)[/tex], the discriminant becomes:
[tex]\[ (b-1)^2 - 4a(c + 5) \][/tex]
4. Set the Discriminant to Zero:
- To have one real solution, set the discriminant equal to zero:
[tex]\[ (b-1)^2 - 4a(c + 5) = 0 \][/tex]
5. Solve for [tex]\( c \)[/tex]:
- Solving the equation for [tex]\( c \)[/tex], we get:
[tex]\[ c = -5 + \frac{(b - 1)^2}{4a} \][/tex]
6. Specific Example:
- Let's find [tex]\( c \)[/tex] for specific values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]. If we set [tex]\( a = 1 \)[/tex] and [tex]\( b = 0 \)[/tex], we can substitute these values into our solution for [tex]\( c \)[/tex]:
[tex]\[ c = -5 + \frac{(0 - 1)^2}{4 \times 1} = -5 + \frac{1}{4} = -5 + 0.25 = -4.75 \][/tex]
- So, for [tex]\( a = 1 \)[/tex] and [tex]\( b = 0 \)[/tex], we have [tex]\( c = -4.75 \)[/tex].
7. Conclusion:
- The parabola given by the equation [tex]\( y = x^2 - 4.75 \)[/tex] will intersect the line [tex]\( y = x - 5 \)[/tex] exactly once, providing us with one real solution.
By following these steps, we can see that the conditions under which a parabola will have exactly one real intersection point with the line [tex]\( y = x - 5 \)[/tex] are given by [tex]\( c = -5 + \frac{(b-1)^2}{4a} \)[/tex], and an example parabola that meets this condition is [tex]\( y = x^2 - 4.75 \)[/tex].
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