Put the line into the parabola and you are left with one quadratic — and the number of solutions it has is the number of times the two graphs meet.
What is the value of k?
In the xy-plane, the graph of the line y = 6x + k and the graph of y = x2 + 2x + 13 intersect at exactly one point, where k is a constant.
Set the two right-hand sides equal and gather everything on one side: x² + 2x + 13 = 6x + k becomes x² − 4x + (13 − k) = 0. Exactly one intersection means exactly one solution, so the discriminant is 0: (−4)² − 4(1)(13 − k) = 16 − 52 + 4k = 0, giving 4k = 36 and k = 9. Check: x² − 4x + 4 = 0 gives x = 2, and both graphs pass through (2, 21). A comes from 4k = −36, a sign slip when 16 − 52 is combined. C forgets to subtract the 6x, working from x² + 2x + (13 − k) = 0 instead. D assumes "one solution" means the constant term is 0, setting 13 − k = 0; that gives x² − 4x = 0, which has two solutions, 0 and 4.
1
Substitute the line into the parabola: two equations become one.
Both equations already say what y is. Set the two expressions for y equal to each other and the y disappears, leaving one quadratic in x.
y = x2 + x − 3 and y = 3x + 5
x² + x − 3 = 3x + 5
the same y, so the two expressions are equal
x² + x − 3 − 3x − 5 = 0
move everything to the side with the x²
x² − 2x − 8 = 0
one quadratic, one variable
(x − 4)(x + 2) = 0, so x = 4 or x = −2
factors of −8 that add to −2
Subtract the whole line, not just part of it
The commonest wreck here is moving the 3x and leaving the 5 behind, or moving the 5 and dropping its sign. Write the subtraction out as a middle line — x² + x − 3 − 3x − 5 — before you collect terms, and the signs take care of themselves.
2
A solution is a point, so every x needs its y.
Put each x back into the line — it is the easier of the two — then check the pair in the parabola.
x = 4: y = 3(4) + 5 = 17
the point (4, 17)
check: 4² + 4 − 3 = 16 + 1 = 17 ✓
the parabola agrees
x = −2: y = 3(−2) + 5 = −1
the point (−2, −1)
check: (−2)² + (−2) − 3 = 4 − 5 = −1 ✓
both graphs pass through both points
the system's solutions: (4, 17) and (−2, −1)
Read the last line of the question before you write an answer. "What is the value of x?" wants a number and often adds a condition such as x > 0 to pick one of the two. "What is the sum of the x-coordinates?" wants 4 + (−2) = 2. "Which is a solution to the system?" wants a whole ordered pair.
3
Two, one, or none — and each has a picture.
A line can cut a parabola twice, graze it once, or miss it entirely. Three parallel lines against y = x² show all three at once.
The picture
The quadratic
The system
line cuts the curve
two real solutions
two solutions (x, y)
line touches the curve
one repeated solution
exactly one solution — the line is tangent
line misses the curve
no real solutions
no solution
The three lines above are parallel, so only the constant changes. Sliding the line down from y = 2x + 3 shrinks the gap between the two crossings until, at y = 2x − 1, they merge into a single point of contact; below that the line has dropped clear of the curve.
4
The discriminant counts the intersections for you.
Once the system is one quadratic ax² + bx + c = 0, the number b² − 4ac decides everything: positive → 2, zero → 1, negative → 0.
You never need the quadratic formula itself here. The question asks how many, not which — and b² − 4ac answers "how many" on its own.
5
“For what value of k is the line tangent?” is the SAT's favourite version.
The stem hides the constant in the line and tells you there is exactly one intersection. That phrase is an instruction: set the discriminant to 0.
y = x2 + 6x + 10 and y = −2x + k
x² + 6x + 10 = −2x + k
equate, keeping k where it is for now
x² + 8x + (10 − k) = 0
add 2x to both sides, subtract k: the constant is 10 − k
8² − 4(1)(10 − k) = 0
exactly one intersection → discriminant 0
64 − 40 + 4k = 0, so 4k = −24 and k = −6
distribute the −4 across both terms
check: x² + 8x + 16 = (x + 4)² = 0 → x = −4
and both graphs give y = 2 there ✓
The constant is 10 − k, not k − 10
k arrives on the right-hand side, so it is subtracted when you move it across. Getting that backwards turns 64 − 4(10 − k) = 0 into 64 − 4(k − 10) = 0, that is 104 − 4k = 0, and answers k = 26. Put k = 26 back into the real discriminant and you get 24 + 4(26) = 128, a positive number: that line cuts the parabola twice rather than touching it. The check line at the end catches this in ten seconds.
The same setup answers the other two phrasings. "Intersect at exactly two points" wants b² − 4ac > 0, which gives a range of values for k rather than a single one; "do not intersect" wants b² − 4ac < 0.
6
When the graph is given, read it instead of solving.
The solutions of the system are exactly the points where the two curves meet, so a graph hands you the answer as a pair of coordinates.
parabola's x-intercepts → its factorsline's y-intercept → its bline's rise over run → its slopemeeting points → the solutions
For a "which system is represented by the graph" item, rebuild each equation from what you can see. A parabola crossing at −2 and 3 with a positive leading coefficient is y = (x + 2)(x − 3); a line through (0, 6) rising two units for every unit right is y = 2x + 6. Then match, rather than testing all four systems point by point.
Desmos is on your screen — use it
Type both equations and the intersections appear as clickable points with their coordinates, which settles any "which is a solution" item outright. For a tangency question, add a slider by typing k as a letter: Desmos offers to create it, and dragging k shows the line rising until it just kisses the curve. Watching the two crossings merge into one is the fastest way to believe the discriminant rule, and on test day it is a reliable check on the value you found by hand.
✓
Check: does the count match the picture?
the cutx² − 2x − 8 = 0disc 36 → (4, 17) and (−2, −1) ✓
the touchx² − 2x + 1 = (x − 1)²disc 0 → one point, at x = 1 ✓
the missx² − 2x + 5 = 0disc −16 → the line never meets it ✓
Remember the flow
Equate · Collect on one side · Count with b² − 4ac · Put each x back for its y
"Exactly one point" always means discriminant zero.
Now you try
Practice problem
Which of the following could be the value of c?
In the xy-plane, the graphs of y = x2 − 5x + 7 and y = x + c intersect at exactly two points, where c is a constant.
Equate and collect: x² − 5x + 7 = x + c gives x² − 6x + (7 − c) = 0. Two intersections means a positive discriminant: (−6)² − 4(1)(7 − c) = 36 − 28 + 4c = 8 + 4c > 0, so c > −2. Of the four values only 1 clears that bar; it gives x² − 6x + 6 = 0 with discriminant 12. C is the boundary itself: c = −2 makes the discriminant 0, so the line is tangent and the graphs meet exactly once, not twice — the single most tempting wrong answer on this question type. A and B are both below −2, giving discriminants −16 and −8, so those lines pass under the parabola and never touch it.
Hint: do not try the four values one at a time. Build the quadratic once, write its discriminant as an expression in c, and then ask which choices make that expression positive — strictly positive, since "exactly two" rules out the tangent case.
