SAT nonlinear systems: find the tangent-line parameter
A phrase such as “the system has exactly one solution” is an instruction about intersections; turn the picture into a single equation before doing any parameter arithmetic.
How do you solve a SAT line-and-parabola system that has exactly one intersection?
Substitute the line into the parabola, collect terms into a quadratic, and set its discriminant to zero to find the parameter that makes the two graphs tangent.
Worked example: a moving line
Problem. The graphs y = x² − 4x + 7 and y = 2x + b meet at exactly one point; find b.
- Set the two expressions for y equal: x² − 4x + 7 = 2x + b.
- Collect terms: x² − 6x + (7 − b) = 0.
- Use one real solution: (−6)² − 4(1)(7 − b) = 0.
- Simplify: 36 − 28 + 4b = 0, giving b = −2.
- The repeated root is x = 3; substituting gives y = 4, so the touching point is (3, 4).
The parameter b is not the y-coordinate of the touching point. Here the line has intercept −2 but touches the parabola at height 4.
Understand which side creates two intersections
The discriminant is 8 + 4b. Therefore b > −2 gives two real intersections, b = −2 gives one, and b < −2 gives none. This inequality check is useful because a wrong sign often still produces a plausible-looking parameter value.
Use Desmos as a check, not a guessing machine
Graph the parabola and the line with b = −2 to inspect the touching point. Then try b = −1 and b = −3 to see the two-intersection and no-intersection cases. Sliders help you understand the change, but derive the exact value algebraically when a parameter is unknown.
Know when this shortcut does not apply
The discriminant test here works because substitution produces a genuine quadratic. It is not a universal test for every nonlinear system: a cubic, a rational equation with excluded inputs, or a square-root equation may need a different method. Always check candidates in the original equations after squaring or clearing denominators.
Try it before you read the answer
1. For what c does y = x² + 2x + 5 meet y = c at exactly one point?
Show explanation
c = 4; rewrite the parabola as y = (x + 1)² + 4, whose minimum is 4.
2. For what k does y = x² and y = 4x + k have exactly one intersection?
Show explanation
k = −4, because x² − 4x − k = 0 has discriminant 16 + 4k; the point is (2, 4).
Practice the decision, not just the arithmetic.
Use VECTOR’s SAT Math practice for Algebra and Functions, then log whether a systems miss came from substitution, rearranging terms, or interpreting the number of solutions.
VECTOR offers a game-based alternative to a conventional practice session. Khan Academy provides official SAT preparation; use the tool that fits the skill you need to learn. No head-to-head score advantage is established.
Sources and scope
The mathematical examples above are independently written, not copied official exam questions. Exam scope and official-practice claims are checked against these sources: