How a line and a parabola can meet — and what it means when they don't.
In 2012, engineers at NASA calculated the exact moment the Mars rover Curiosity would intersect the Martian surface. They had two equations: one describing the rover's curved descent path, and one describing the ground. Finding where those two equations were equal — that is solving a system of a quadratic and a linear equation. The same idea shows up whenever a straight-line path crosses a curved one.
A system of a quadratic and a linear equation asks: where do a parabola and a line share the same point? The answer comes from setting the two expressions equal and solving. You already know how to solve quadratic equations by factoring or the quadratic formula. The only new step here is the setup.
Say you have the system:
y=x2−2x−3
y=x+1
Both expressions equal y, so they equal each other. That gives you one equation in one variable.
x2−2x−3=x+1
Move everything to one side:
x2−3x−4=0
Factor:
(x−4)(x+1)=0
So x=4 or x=−1. Plug each back into the simpler equation, y=x+1, to find the y-coordinates. When x=4, y=5. When x=−1, y=0. The two intersection points are (4,5) and (−1,0).
That process — substitute, rearrange, solve the resulting quadratic, find y — is the full algebraic method. Here it is with every step shown.
Solving y = x² + x − 2 and y = 2x + 1
x2+x−2=2x+1
Both sides equal y, so set the right-hand sides equal to each other.
x2+x−2x−2−1=0
Move everything from the right side to the left. Subtract 2x and subtract 1 from both sides.
x2−x−3=0
Combine like terms: x - 2x = -x, and -2 - 1 = -3.
x=2(1)1±(−1)2−4(1)(−3)
This doesn't factor nicely, so use the quadratic formula. Here a = 1, b = -1, c = -3.
x=21±1+12
Under the radical: (-1)² = 1, and -4(1)(-3) = 12.
x=21±13
Simplify under the radical. 13 has no perfect square factors, so this is the simplest form.
x≈2.30orx≈−1.30
Two decimal approximations. You need both — the system has two solutions.
y≈2(2.30)+1=5.60andy≈2(−1.30)+1=−1.60
Plug each x back into y = 2x + 1. Use the linear equation — it's easier arithmetic.
(2.30,5.60)and(−1.30,−1.60)✓
State both intersection points. On the Regents, round to the nearest hundredth unless the problem asks otherwise.
The graphical picture makes the algebra concrete. The parabola and the line cross at exactly the two points the algebra found. Here is what that looks like.
Interactive graph — scroll to zoom, drag to pan
A line and a parabola can intersect in three ways. Two solutions: the line cuts through the parabola at two points, as in the examples above. One solution: the line is tangent to the parabola — it just grazes it at exactly one point. Zero solutions: the line misses the parabola entirely. The discriminant, b2−4ac, tells you which case you are in before you finish solving. Positive discriminant means two intersection points. Zero discriminant means exactly one. Negative discriminant means none — no real solutions exist.
Now a second worked example where the solution comes out to clean integers.
Solving y = x² − 4x + 5 and y = −x + 5
x2−4x+5=−x+5
Set the two expressions equal. Both equal y.
x2−4x+x+5−5=0
Add x to both sides and subtract 5 from both sides.
x2−3x=0
Combine like terms. Notice there is no constant term left.
x(x−3)=0
Factor out x. One factor is just x itself.
x=0orx=3
Set each factor equal to zero.
y=−(0)+5=5andy=−(3)+5=2
Substitute both x values into y = -x + 5.
(0,5)and(3,2)✓
Two clean intersection points. Always write them as coordinate pairs.
Practice Questions
y=x2−3x+2 and y=x−1
y=x2+2x−8 and y=2x
y=x2+4 and y=x+2
Regents Corner
On Part II and Part III of the Algebra I Regents, linear-quadratic system problems often ask you to solve both algebraically and graphically, and they award separate points for each method. Skipping one earns partial credit at best. For the algebraic solution, you must show the substitution step, the rearranged quadratic set equal to zero, and both solutions as ordered pairs — not just the x values. For the graphical solution, your sketch must show both curves labeled, and you must mark the intersection points with their coordinates.
After solving and finding x = 1 and x = 3, students write (1, 3) as a single answer — treating the two x-values as one coordinate pair. These are two separate x-values that need two separate y-values. You must substitute each x into one of the original equations to get y, then write two complete ordered pairs.