Algebra 1 · A.REI.C.7

Systems with a curve

A straight line and a parabola are drawn on the same axes. Where do they meet, and how many times can they?

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Solving systems by substitutionMath 8 Quadratic expressions and parabolasGrade 9

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When you reach for it

A system of two equations in x and y asks for every pair (x, y) that makes both equations true at once. When both are lines there is at most one such pair, found by graphing, substitution or elimination. This lesson takes the next case: one equation is a line and the other is a parabola, such as

y = x^2 - 2x - 3 \quad\text{and}\quad y = x + 1

A solution is still a pair that makes both equations true, so it is still a point on both graphs. What changes is how many there can be. A line and a parabola can cross twice, meet once, or miss each other, and the method has to find every meeting point, not stop at the first.

Reach for it whenever two conditions hold at once and one of them is curved: where a straight beam strikes a curved surface, or simply where two graphs meet. Reading meeting points off a graph to solve f(x) = g(x) is the same picture; this lesson adds the algebra that gives the points exactly.

The procedure

What a solution looks like

A point lies on a graph exactly when its coordinates make the graph’s equation true. So a pair satisfying both equations is a point lying on both graphs, and the solutions of the system are the meeting points of the line and the parabola.

Figure 1The parabola y = x^2 - 4x + 5 against four lines. A line can cross it twice, touch it once, meet it once by being vertical, or miss it.

A line that is not vertical can cross the parabola at two points, touch it at one point and stay on one side of it, or miss it. A vertical line x = h meets a parabola that opens up or down exactly once, because the parabola has exactly one height above each x. There is never a third meeting point; the substitution below shows why.

Solving by substitution

Take the system above. Both equations say what y is, and at a meeting point the two y-values are the same number, so the two expressions for y are equal:

x^2 - 2x - 3 = x + 1

This is substitution: the line’s expression for y has replaced y in the parabola’s equation. The result is one equation in one unknown, and it is quadratic.

Collect everything on one side. Subtract x + 1 from both sides to reach the standard form ax^2 + bx + c = 0:

x^2 - 3x - 4 = 0

The zero matters: the zero product property splits only a product that equals zero.

Solve the quadratic by whichever method fits: factoring, square roots when there is no x term, or the quadratic formula when nothing factors. Here (x - 4)(x + 1) = 0, so x = 4 or x = -1.

Find each y from the line. Each x is the first coordinate of a meeting point, and that point lies on the line, so the line gives its height: x = 4 gives y = 5, and x = -1 gives y = 0. The parabola would give the same heights, since the point is on both graphs, but the line’s arithmetic is shorter.

Check in both equations. For (4, 5): 16 - 8 - 3 = 5 and 4 + 1 = 5. For (-1, 0): 1 + 2 - 3 = 0 and -1 + 1 = 0. Every step above can be undone, so no false solution creeps in; the check catches arithmetic slips.

The solutions are the two points (-1, 0) and (4, 5), each a pair.

Figure 2The system y = x^2 - 2x - 3, y = x + 1. Its two solutions are the two crossing points.

Why never three. A line y = mx + k has no x^2 term, so nothing cancels the parabola’s ax^2 when the sides are collected. The combined equation is a quadratic, which has at most two solutions, and each solution x gives one point. A vertical line skips the quadratic: putting x = h into the parabola’s equation gives one y.

When the line is not solved for y. A line written as 3x + y = 7 does not yet say what y is. Rearrange it first, doing the same thing to both sides: y = 7 - 3x. Then substitute (Example 2).

Graphing first. A sketch of both graphs shows how many meeting points there are and roughly where. It rarely gives them exactly: a crossing at \displaystyle x = 1 + \sqrt{3} looks like “about 2.7”. The graph is the estimate, the algebra is the answer, and each checks the other (Example 3).

Counting without solving

The meeting points match the real solutions of the combined equation ax^2 + bx + c = 0, so its discriminant b^2 - 4ac counts them before anything is solved:

b^2 - 4ac	Meeting points	The picture
positive	two	the line crosses the parabola
zero	one	the line touches it
negative	none	the line misses it

This is the discriminant doing its usual job, and the discriminant and the graph shows the same count for one parabola and the x-axis. Use the combined equation’s discriminant: the parabola’s own says only how it meets the line y = 0.

Worked examples

Example 1

A downward parabola

Solve the system y = -x^2 + 2x + 8 and y = x + 2.

Solution

Both equations give y, so set the expressions equal and collect on the left:

\begin{aligned} -x^2 + 2x + 8 &= x + 2 \\ -x^2 + x + 6 &= 0 \end{aligned}

Multiply both sides by -1 so the x^2 term is positive; this does not change the solutions, since 0 \times (-1) = 0. Then factor:

\begin{aligned} x^2 - x - 6 &= 0 \\ (x - 3)(x + 2) &= 0 \end{aligned}

So x = 3 or x = -2. The line gives the heights: y = 3 + 2 = 5 and y = -2 + 2 = 0.

Check in the parabola: -9 + 6 + 8 = 5 and -4 - 4 + 8 = 0. The solutions are (3, 5) and (-2, 0). The second lies on the x-axis: both graphs cross the axis at x = -2.

Example 2

A line in standard form

Solve the system 3x + y = 7 and y = x^2 - 2x + 1.

Solution

The line does not say what y is yet. Subtract 3x from both sides: y = 7 - 3x. Now set the expressions for y equal, then add 3x to both sides and subtract 7:

\begin{aligned} x^2 - 2x + 1 &= 7 - 3x \\ x^2 + x - 6 &= 0 \end{aligned}

This factors as (x + 3)(x - 2) = 0, so x = -3 or x = 2. From y = 7 - 3x: x = -3 gives y = 16, and x = 2 gives y = 1.

Check against the line as it was first given, so that a slip in rearranging would show: 3(-3) + 16 = 7 and 9 + 6 + 1 = 16; 3(2) + 1 = 7 and 4 - 4 + 1 = 1. The solutions are (-3, 16) and (2, 1).

Example 3

Reading a graph, then solving exactly

The graphs of y = x^2 - 1 and y = 2x + 1 are drawn below. Estimate the solutions of the system from the graph, then find them exactly.

Figure 3Example 3. The graph shows roughly where the crossings are; the algebra says exactly.

Solution

The graphs cross twice, near (2.7, 6.5) and near (-0.7, -0.5). Neither crossing sits on a grid point, so the graph cannot give exact values.

Set the expressions equal and collect:

\begin{aligned} x^2 - 1 &= 2x + 1 \\ x^2 - 2x - 2 &= 0 \end{aligned}

No two integers multiply to -2 and add to -2, so use the quadratic formula with a = 1, b = -2, c = -2:

x = \frac{2 \pm \sqrt{4 + 8}}{2} = \frac{2 \pm 2\sqrt{3}}{2} = 1 \pm \sqrt{3}

The line gives \displaystyle y = 2(1 \pm \sqrt{3}) + 1 = 3 \pm 2\sqrt{3}. The signs stay paired: the point with \displaystyle x = 1 + \sqrt{3} has \displaystyle y = 3 + 2\sqrt{3}. Mixing them gives a point on neither graph. So the solutions are

\begin{aligned} &\left(1 + \sqrt{3},\ 3 + 2\sqrt{3}\right) \quad\text{and} \\ &\left(1 - \sqrt{3},\ 3 - 2\sqrt{3}\right) \end{aligned}

which are about (2.7, 6.5) and (-0.7, -0.5), matching the estimate. Each y came from the line, so the line holds. Check both in the parabola, signs matched: \displaystyle (1 \pm \sqrt{3})^2 - 1 = 1 \pm 2\sqrt{3} + 3 - 1 = 3 \pm 2\sqrt{3}.

Example 4

Counting the meeting points

Without solving, decide how many points each pair of graphs shares, and find the point when there is only one.

(a) y = x^2 - 4x + 7 and y = 2x - 2 \qquad (b) y = 2x^2 + 3x and y = x - 1

Solution

In (a) to (c), set the expressions equal, collect, and read the discriminant.

(a) x^2 - 6x + 9 = 0, and b^2 - 4ac = 36 - 36 = 0: one point, where the line touches. Since x^2 - 6x + 9 = (x - 3)^2, the point has x = 3 and y = 2(3) - 2 = 4. It is (3, 4).

(b) 2x^2 + 2x + 1 = 0, and b^2 - 4ac = 4 - 8 = -4: no points. The line passes below the parabola.

(c) -x^2 + 2x + 3 = 0, with a = -1, b = 2, c = 3, so b^2 - 4ac = 4 + 12 = 16: two points. There is no need to make a positive first, since multiplying by -1 leaves b^2 - 4ac unchanged.

(d) The line is vertical, so substitute x = 5: y = 25 - 20 + 7 = 12. One point, (5, 12), with no discriminant needed. The lines in (a) and (d) each meet the same parabola once, but differently: the touching line stays on one side of the curve, while the vertical line passes into the cup.

Example 5

A laser across a dish

A dish has a parabolic cross-section. With the origin at its lowest point and lengths in centimeters, its inner surface is \displaystyle y = \frac{x^2}{40} for -40 \le x \le 40, so its rim is at (-40, 40) and (40, 40). Lucasta fixes a laser at the right rim and aims it across the dish along \displaystyle y = \frac{1}{2}x + 20. Where does the beam strike the inside of the dish?

Figure 4Example 5. The beam leaves the rim at (40, 40) and strikes the dish at (-20, 10).

Solution

The point struck is on both the curve and the beam’s line, so solve the system. Set the expressions equal and multiply both sides by 40 to clear the fractions:

\begin{aligned} \frac{x^2}{40} &= \frac{1}{2}x + 20 \\ x^2 &= 20x + 800 \end{aligned}

Collect and factor:

\begin{aligned} x^2 - 20x - 800 &= 0 \\ (x - 40)(x + 20) &= 0 \end{aligned}

The root x = 40 is the rim, where the laser sits: that point is on both graphs too, so the factor x - 40 had to appear. The other root, x = -20, is inside -40 \le x \le 40, and the line gives y = -10 + 20 = 10. Check on the dish: \displaystyle \frac{400}{40} = 10.

The beam strikes the dish at (-20, 10): 20 cm to the left of the center and 10 cm above the lowest point.

Example 6

Two parabolas

Solve the system y = x^2 - 3 and y = -x^2 + 2x + 1.

Solution

Set the expressions equal, collect, and divide both sides by 2:

\begin{aligned} x^2 - 3 &= -x^2 + 2x + 1 \\ x^2 - x - 2 &= 0 \end{aligned}

So (x - 2)(x + 1) = 0, and x = 2 or x = -1. There is no line to take y from, so use one parabola and check in the other. x = 2 gives y = 4 - 3 = 1, and -4 + 4 + 1 = 1. x = -1 gives y = 1 - 3 = -2, and -1 - 2 + 1 = -2. The solutions are (2, 1) and (-1, -2).

There is a second route. The x^2 terms are opposites, so adding the two equations, as in elimination, removes them: 2y = 2x - 2, that is, y = x - 1. Every common point of the two parabolas satisfies this, since adding two true equations gives a true one, and indeed 1 = 2 - 1 and -2 = -1 - 1. The step also runs backwards: twice the line’s equation minus one parabola’s gives the other’s. So the two parabolas meet exactly where the line y = x - 1 meets either one of them.

Boundaries

Real numbers only. A negative discriminant means no meeting point. The combined equation then has complex solutions, but they are not points on either graph.

The line gives y only if it is not vertical. Ax + By = C can be solved for y only when B \ne 0. When B = 0 the line is x = h: substitute h into the parabola.

At most two points, with one exception. A line meets y = ax^2 + bx + c (with a \ne 0) at most twice. Two parabolas usually do too, but if their x^2 coefficients are equal the combined equation is linear, so there is at most one point: y = x^2 and y = x^2 + 2x meet only at (0, 0). Two identical parabolas share every point.

Coordinates come in pairs. Each x goes with its own y; with \pm answers, the signs stay matched.

In context, a solution must also fit the setting. In Example 5 the root x = 40 was where the laser sat, not where it struck, and any root outside -40 \le x \le 40 would have been off the dish.

Where this goes next

The meeting points of two graphs are where one stops being above the other, so the area between curves starts by finding them: the region is split at exactly these x-values. The same substitution later finds where a line meets a circle such as x^2 + y^2 = 25, where the counts are again two, one or none.

Exercises

Points on both graphs

Which of (0, -3), (1, -2) and (2, 1) are solutions of the system y = x^2 - 3, y = 2x - 3?

Solve by substitution

y = x^2 and y = x + 6
y = x^2 - 4x + 1 and y = x - 3
y = 2x^2 - 3x and y = x + 6
y = -x^2 + 5 and y = 2x + 2
y = x^2 + 2x - 1 and y = 2x + 8

Lines in other forms

x + y = 3 and y = x^2 - 3x
4x - 2y = 6 and y = x^2 - 2x - 1
x = -1 and y = 3x^2 - x + 2

Exact answers and estimates

Solve y = x^2 + 1, y = 4x exactly, then give each point to one decimal place.
The graphs of y = x^2 - 2 and y = -x seem to cross at (-2, 2) and (1, -1). Confirm by solving the system.

How many meeting points

How many points do y = x^2 + 3x + 5 and y = x + 4 share? Find them.
How many points do y = x^2 - x + 3 and y = 2x - 1 share?
For which value of c does the line y = -2x + c touch the parabola y = x^2 + 2x + 3, and where? For which values of c does it cross the parabola twice?

Two parabolas, and a dish

Solve the system y = x^2 + 1 and y = -x^2 + 9.
Fyodor fixes a laser at the left rim, (-40, 40), of the dish \displaystyle y = \frac{x^2}{40} from Example 5, and aims it along \displaystyle y = -\frac{1}{4}x + 30. Where does the beam strike the dish?

Reasoning

Explain why a line and the parabola y = ax^2 + bx + c, with a \ne 0, can never meet at three points.
Show that every line y = mx + 1 meets the parabola y = x^2 at exactly two points, whatever the value of m. Explain why this makes sense from the picture.

Answers

(0, -3) and (2, 1). Both satisfy both equations: 0 - 3 = -3 = 0 - 3, and 4 - 3 = 1 = 4 - 3. (1, -2) is on the parabola, 1 - 3 = -2, but not the line, 2 - 3 = -1.
(3, 9) and (-2, 4). x^2 - x - 6 = 0 gives (x - 3)(x + 2) = 0.
(1, -2) and (4, 1). x^2 - 5x + 4 = 0 gives (x - 1)(x - 4) = 0; the line gives each y.
(3, 9) and (-1, 5). 2x^2 - 4x - 6 = 0; divide by 2: (x - 3)(x + 1) = 0.
(1, 4) and (-3, -4). -x^2 - 2x + 3 = 0; multiply by -1: (x - 1)(x + 3) = 0.
(3, 14) and (-3, 2). The 2x terms cancel: x^2 = 9, so x = \pm 3 by square roots.
(3, 0) and (-1, 4). y = 3 - x, so x^2 - 3x = 3 - x, x^2 - 2x - 3 = 0, (x - 3)(x + 1) = 0.
\displaystyle \left(2 + \sqrt{2},\ 1 + 2\sqrt{2}\right) and \displaystyle \left(2 - \sqrt{2},\ 1 - 2\sqrt{2}\right). The line is y = 2x - 3, so x^2 - 4x + 2 = 0 and \displaystyle x = \frac{4 \pm \sqrt{8}}{2} = 2 \pm \sqrt{2}.
(-1, 6), one point. Substitute x = -1: y = 3 + 1 + 2 = 6.
\displaystyle \left(2 + \sqrt{3},\ 8 + 4\sqrt{3}\right) \approx (3.7, 14.9) and \displaystyle \left(2 - \sqrt{3},\ 8 - 4\sqrt{3}\right) \approx (0.3, 1.1). x^2 - 4x + 1 = 0 gives \displaystyle x = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}, and y = 4x.
Confirmed: (-2, 2) and (1, -1). x^2 - 2 = -x gives x^2 + x - 2 = 0, (x + 2)(x - 1) = 0, and y = -x.
One point, (-1, 3), where the line touches. x^2 + 2x + 1 = 0 has b^2 - 4ac = 4 - 4 = 0, and (x + 1)^2 = 0 gives x = -1.
None. x^2 - 3x + 4 = 0 has b^2 - 4ac = 9 - 16 = -7.
It touches when c = -1, at (-2, 3); it crosses twice when c > -1. The combined equation is x^2 + 4x + (3 - c) = 0, with b^2 - 4ac = 16 - 4(3 - c) = 4 + 4c. That is 0 at c = -1, where (x + 2)^2 = 0, and positive when c > -1.
(2, 5) and (-2, 5). x^2 + 1 = -x^2 + 9 gives 2x^2 = 8, so x = \pm 2.
At (30, 22.5). \displaystyle \frac{x^2}{40} = -\frac{1}{4}x + 30 gives x^2 + 10x - 1200 = 0, that is (x + 40)(x - 30) = 0. The root x = -40 is the laser’s own position.
Substituting y = mx + k gives ax^2 + bx + c = mx + k, and collecting leaves ax^2 + (b - m)x + (c - k) = 0. The line has no x^2 term, so a \ne 0 survives and this is a quadratic, with at most two solutions; each gives one point. A vertical line x = h meets the parabola once, at y = ah^2 + bh + c.
The combined equation is x^2 - mx - 1 = 0, with b^2 - 4ac = m^2 + 4, which is positive for every m, so there are always two points. In the picture, (0, 1) lies inside the cup, above the vertex (0, 0). A non-vertical line through a point inside the cup has to leave it on both sides, crossing the parabola once on each.