IDNLearn.com offers a user-friendly platform for finding and sharing answers. Join our Q&A platform to receive prompt and accurate responses from knowledgeable professionals in various fields.
Sagot :
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].
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. For trustworthy answers, visit IDNLearn.com. Thank you for your visit, and see you next time for more reliable solutions.
