Question:The parabola P has equation \(\mathrm{y = (x - 4)^2 + 1}\).The line L has slope 1 and passes through...

1. TRANSLATE the line information into an equation

• Given information:
  • Line L has slope \(\mathrm{m = 1}\)
  • Line L passes through point \(\mathrm{(t, 2t + 3)}\)
  • Need to find equation of this line

• Using point-slope form: \(\mathrm{y - y_1 = m(x - x_1)}\)
  • \(\mathrm{y - (2t + 3) = 1(x - t)}\)
  • \(\mathrm{y - 2t - 3 = x - t}\)
  • \(\mathrm{y = x + t + 3}\)

2. INFER what "exactly one intersection" means mathematically

• For the graphs to intersect at exactly one point, we need to solve:
  Parabola equation = Line equation
  \(\mathrm{(x - 4)^2 + 1 = x + t + 3}\)

• This will create a quadratic equation in x, and for exactly one solution, the discriminant must equal zero.

3. SIMPLIFY to create the quadratic equation

• Start with: \(\mathrm{(x - 4)^2 + 1 = x + t + 3}\)
• Expand the left side: \(\mathrm{x^2 - 8x + 16 + 1 = x + t + 3}\)
• Collect all terms: \(\mathrm{x^2 - 8x + 17 - x - t - 3 = 0}\)
• Final quadratic: \(\mathrm{x^2 - 9x + (14 - t) = 0}\)

4. APPLY the discriminant condition

• For quadratic \(\mathrm{ax^2 + bx + c = 0}\), discriminant \(\mathrm{D = b^2 - 4ac}\)
• Here: \(\mathrm{a = 1, b = -9, c = 14 - t}\)
• \(\mathrm{D = (-9)^2 - 4(1)(14 - t)}\)
  \(\mathrm{= 81 - 56 + 4t}\)
  \(\mathrm{= 25 + 4t}\)

5. SIMPLIFY to solve for t

• Set discriminant equal to zero: \(\mathrm{25 + 4t = 0}\)
• Solve: \(\mathrm{4t = -25}\)
• \(\mathrm{t = -\frac{25}{4}}\)

Answer: \(\mathrm{-\frac{25}{4}}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students struggle to convert "passes through point \(\mathrm{(t, 2t + 3)}\)" into a usable line equation, especially when the point coordinates contain the unknown parameter t.

Many students either:

• Forget to use point-slope form entirely
• Make algebraic errors when substituting the point into point-slope form
• Get confused by having t appear in both the x and y coordinates

This leads to an incorrect line equation, making the entire subsequent solution wrong and causing them to guess randomly.

Second Most Common Error:

Poor INFER reasoning: Students don't connect "exactly one intersection point" with the discriminant condition.

They might attempt to solve the system algebraically by substitution but fail to recognize that they need the discriminant to equal zero for a unique solution. Instead, they might try to solve for specific intersection coordinates, leading to confusion about how to use the parameter t.

This causes them to get stuck and abandon the systematic approach.

The Bottom Line:

This problem requires strong algebraic manipulation skills combined with conceptual understanding of when parabolas and lines are tangent (touch at exactly one point). The parameter t adds complexity that demands careful attention to both translation and logical reasoning about intersection conditions.