Nonlinear Systems and Curve Intersections, precalculus lesson

A straight road cuts across a circular track. A thrown ball follows a parabola while a ramp is a straight line. Where do they meet? Two different lines meet at most once, but curves can meet two, three, four or even more times, and calculus later needs every one of those points, for example as the endpoints when finding the area between two curves.

Key terms

Intersection points

Definition A nonlinear system is a system of equations in which at least one equation is not linear. A point $(a, b)$ is an intersection point of the graphs of two equations in $x$ and $y$ exactly when it satisfies both equations. So the real solutions of the system are exactly the intersection points of the graphs. A nonlinear system can have no solution, any finite number of solutions, or infinitely many.

In plain words Every point where the two graphs cross or touch is a solution, and every real solution is such a point. Curves can meet more often than lines can.

Example The curves $y = x^3$ and $y = x$ meet where $x^3 = x$ , that is where

x^3 - x = x(x - 1)(x + 1) = 0.

So there are three intersection points, $(-1, -1)$ , $(0, 0)$ and $(1, 1)$ , something no pair of distinct lines can produce.

Substitution

Definition To solve by substitution, solve one equation for one variable, or for an expression such as $x^2$ , and replace it in the other equation. Choose the variable or expression so that a single equation in one variable remains. Solve it, then for each root find the other coordinate from an equation that determines it uniquely.

In plain words Turn two equations in two unknowns into one equation in one unknown. A linear equation is usually the easiest one to solve for a variable.

Example Two numbers have sum $5$ and product $6$ . From $x + y = 5$ and $xy = 6$ , substitute $y = 5 - x$ into the second equation.

\begin{gathered} x(5 - x) = 6, \\[0.4em] x^2 - 5x + 6 = 0, \\[0.4em] (x - 2)(x - 3) = 0. \end{gathered}

So $x = 2$ or $x = 3$ , and $y = 5 - x$ gives the points $(2, 3)$ and $(3, 2)$ , where the line meets the hyperbola $xy = 6$ .

Elimination

Definition Elimination replaces one equation by its sum with a constant multiple of the other, chosen so that a variable, or a whole term such as $x^2$ or $y^2$ , cancels. As in linear systems, this step and multiplying an equation by a nonzero constant do not change the solution set. It works well when both equations contain the same nonlinear terms.

In plain words If both equations contain $x^2$ and $y^2$ , add or subtract them so that one of those terms drops out, just as you would with $x$ and $y$ in a linear system.

Example Solve $x^2 + y^2 = 13$ and $x^2 - y^2 = 5$ . Adding gives $2x^2 = 18$ , so $x = 3$ or $x = -3$ . Subtracting gives $2y^2 = 8$ , so $y = 2$ or $y = -2$ . Only squares appear in the equations, so every combination works, and the circle meets the hyperbola in four points, $(3, 2)$ , $(3, -2)$ , $(-3, 2)$ and $(-3, -2)$ .

Line and parabola

Definition Let a parabola be $y = ax^2 + bx + c$ with $a \ne 0$ , and let a nonvertical line be $y = mx + k$ . Substitution gives the quadratic equation

ax^2 + (b - m)x + (c - k) = 0,

whose real roots are the $x$ -coordinates of the intersection points. With discriminant

D = (b - m)^2 - 4a(c - k),

there are two intersection points if $D > 0$ , one if $D = 0$ and none if $D < 0$ . A vertical line $x = h$ meets the parabola exactly once, at the point $(h, ah^2 + bh + c)$ .

In plain words Set the two expressions for $y$ equal and solve the quadratic. Its discriminant tells you in advance whether the line cuts through the parabola, touches it or misses it.

Example Where does $y = x + 1$ meet $y = x^2 - 4x + 5$ ? Setting the right sides equal gives

\begin{gathered} x^2 - 5x + 4 = 0, \\[0.4em] (x - 1)(x - 4) = 0. \end{gathered}

So $x = 1$ or $x = 4$ , and the line gives the points $(1, 2)$ and $(4, 5)$ . Both lie on the parabola as well, since $1 - 4 + 5 = 2$ and $16 - 16 + 5 = 5$ .

Circle and line

Definition Let a circle have center $(h, k)$ and radius $r > 0$ , so its equation is $(x - h)^2 + (y - k)^2 = r^2$ , and let a line be $Ax + By + C = 0$ with $A$ and $B$ not both zero. The distance from the center to the line is

d = \frac{\lvert Ah + Bk + C \rvert}{\sqrt{A^2 + B^2}}.

The line meets the circle in two points if $d < r$ , in exactly one point if $d = r$ and in no point if $d > r$ . To find the points, solve the line equation for one variable and substitute into the circle equation.

In plain words A line passes through a circle, touches it or misses it, depending on how close it comes to the center. Substitution gives a quadratic whose real roots locate the points.

Example Where does the line $y = x + 1$ meet the circle $x^2 + y^2 = 25$ ? Substituting,

\begin{gathered} x^2 + (x + 1)^2 = 25, \\[0.4em] 2x^2 + 2x - 24 = 0, \\[0.4em] x^2 + x - 12 = 0. \end{gathered}

So $(x + 4)(x - 3) = 0$ , which gives $x = 3$ or $x = -4$ , and the line gives the points $(3, 4)$ and $(-4, -3)$ . Written as $x - y + 1 = 0$ , the line is at distance $\frac{1}{\sqrt{2}} \approx 0.71$ from the origin, less than the radius $5$ , which agrees with two points.

Tangency

Definition A line is tangent to a circle when it meets the circle in exactly one point, and then it is perpendicular to the radius at that point. A nonvertical line is tangent to the parabola $y = ax^2 + bx + c$ when it meets the parabola in exactly one point, which happens exactly when the quadratic from substitution has a double root, that is, when $D = 0$ . A vertical line also meets the parabola once, but it crosses the curve instead of touching it, so it is not tangent.

In plain words Tangency is the boundary case between two intersection points and none. In the algebra, the two roots of the quadratic merge into one repeated root.

Example For which $k$ is the line $y = 2x + k$ tangent to $y = x^2$ ? Substitution gives $x^2 - 2x - k = 0$ , with discriminant

D = (-2)^2 - 4 \cdot 1 \cdot (-k) = 4 + 4k.

Tangency needs $D = 0$ , so $k = -1$ . The quadratic is then $x^2 - 2x + 1 = (x - 1)^2$ , with the double root $x = 1$ , and the line $y = 2x - 1$ touches the parabola at $(1, 1)$ .

Checking every equation

Definition A candidate pair is a solution of the system only if it satisfies every original equation. When you find the second coordinate from an equation that allows several values, such as $y^2 = 16$ , some of the resulting pairs may satisfy one equation and fail the other. So substitute every candidate into every original equation.

In plain words A solution has to lie on both curves. Put each candidate back into both equations before you call it an answer.

Example In the circle and line example, one root was $x = 3$ . Taking $y$ from the circle $x^2 + y^2 = 25$ gives $y^2 = 16$ , so $y = 4$ or $y = -4$ . The point $(3, -4)$ lies on the circle, but $-4 \ne 3 + 1$ , so it is not on the line $y = x + 1$ . Only $(3, 4)$ is an intersection point, which is why the line, not the circle, should supply the second coordinate.

Extraneous solutions

Definition An extraneous solution is a candidate that the solving process produces but that does not satisfy the original system. It can appear after a step that is not reversible, such as squaring both sides of an equation or multiplying by an expression that can be zero. Squaring keeps every real solution but may add new ones, so each candidate must be checked in the original equations, as with radical equations.

In plain words Squaring forgets the difference between $u = v$ and $u = -v$ , so the squared equation can have extra roots. They look like real answers until you test them.

Example Find where $y = \sqrt{x}$ meets $y = x - 2$ . Setting $\sqrt{x} = x - 2$ and squaring gives

\begin{gathered} x = x^2 - 4x + 4, \\[0.4em] x^2 - 5x + 4 = 0, \\[0.4em] x = 1 \text{ or } x = 4. \end{gathered}

At $x = 4$ both curves give $y = 2$ , so $(4, 2)$ is an intersection point. At $x = 1$ the root curve gives $\sqrt{1} = 1$ but the line gives $1 - 2 = -1$ , so $x = 1$ is extraneous. It solves $-\sqrt{x} = x - 2$ instead.

Check yourself

Find all intersection points of $y = x^2 - 2x - 3$ and $y = x + 1$ .

Show the answer

Setting the right sides equal gives $x^2 - 3x - 4 = 0$ , or $(x - 4)(x + 1) = 0$ . So $x = 4$ or $x = -1$ , and the line gives $(4, 5)$ and $(-1, 0)$ . Both check in the parabola, since $16 - 8 - 3 = 5$ and $1 + 2 - 3 = 0$ .

For which values of $b$ does the line $y = x + b$ meet the circle $x^2 + y^2 = 8$ in exactly one point? Find that point for each value.

Show the answer

Substitution gives $2x^2 + 2bx + b^2 - 8 = 0$ , with discriminant

\begin{aligned} D &= (2b)^2 - 4 \cdot 2 \cdot (b^2 - 8) \\[0.4em] &= 64 - 4b^2. \end{aligned}

Tangency needs $D = 0$ , so $b = 4$ or $b = -4$ . The double root is $x = -\frac{b}{2}$ , which gives the point $(-2, 2)$ for $b = 4$ and the point $(2, -2)$ for $b = -4$ . Both points satisfy $x^2 + y^2 = 8$ .

Solve the system $y = \sqrt{x + 3}$ and $y = x + 1$ .

Show the answer

Squaring $\sqrt{x + 3} = x + 1$ gives $x + 3 = x^2 + 2x + 1$ , or $x^2 + x - 2 = 0$ , so $x = 1$ or $x = -2$ . At $x = 1$ both equations give $y = 2$ . At $x = -2$ the root gives $1$ but the line gives $-1$ , so $x = -2$ is extraneous. The only solution is $(1, 2)$ .

Nonlinear Systems and Curve Intersections

Key terms

Intersection points

Substitution

Elimination

Line and parabola

Circle and line

Tangency

Checking every equation

Extraneous solutions

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus