Math Core

Where does a thrown ball land on a sloped hill? Where does a straight road cross a curved one? Each question asks where a line meets a curve, which means solving a system with one linear equation and one quadratic equation. Substitution turns the system into a single quadratic, and everything you've learned in this unit applies.

How many solutions?

A line and a parabola can meet in two points, touch at exactly one point, or miss each other entirely. (Two different lines can never meet twice, which is why this is new.)

The parabola y = x² - 4x + 3 crosses y = x - 1 at two points, touches y = -4x + 3 only at (0, 3), and never meets y = -x - 2.Open in grapher →

Solving a linear-quadratic system

Solve the linear equation for one variable (usually $y$ ), unless it is already solved.
Substitute into the quadratic equation to get a quadratic equation in one variable.
Solve that quadratic. Its real solutions are the x-coordinates of the intersection points.
Substitute each x-value into the linear equation to find the matching y-value.

The discriminant of the quadratic in step 2 tells you the number of intersection points: two if $D > 0$ , one (the line is tangent) if $D = 0$ , none if $D < 0$ .

Two intersection points

Worked example: Substitution

Solve the system $y = x^2 - 4x + 3$ and $y = x - 1$ .

Both expressions equal $y$ , so set them equal:

\begin{aligned} x^2 - 4x + 3 &= x - 1 \\ x^2 - 5x + 4 &= 0 \\ (x - 1)(x - 4) &= 0 \end{aligned}

So $x = 1$ or $x = 4$ . Use the line to find $y$ : when $x = 1$ , $y = 0$ ; when $x = 4$ , $y = 3$ . The solutions are $(1, 0)$ and $(4, 3)$ .

Check $(4, 3)$ in the parabola: $16 - 16 + 3 = 3$ . It works.

Common mistake

Each solution of a system is a point, an ordered pair. Stopping at $x = 1$ and $x = 4$ is only half the answer. Also, pair each x-value with its own y-value: $(1, 3)$ and $(4, 0)$ are not solutions. Finding $y$ from the linear equation is easiest and avoids pairing mistakes.

One point or none

Worked example: A tangent line

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

x^2 + 2 = 2x + 1 \quad\Longrightarrow\quad x^2 - 2x + 1 = 0 \quad\Longrightarrow\quad (x - 1)^2 = 0.

The only solution is $x = 1$ , and then $y = 2(1) + 1 = 3$ . The system has one solution, $(1, 3)$ : the line just touches the parabola there. Such a line is called tangent to the parabola.

Worked example: No intersection

How many solutions does the system $y = -x^2 + 1$ and $y = x + 3$ have?

-x^2 + 1 = x + 3 \quad\Longrightarrow\quad x^2 + x + 2 = 0.

The discriminant is $1 - 8 = -7 < 0$ . There are no real solutions, so the line and the parabola never meet. The quadratic does have complex solutions, but they don't correspond to points on the graph, because points in the coordinate plane have real coordinates.

Circles and other quadratic equations

The same method works when the quadratic equation is a circle, such as $x^2 + y^2 = 25$ (the circle of radius $5$ centered at the origin). A line can cross a circle twice, touch it once, or miss it.

Worked example: A line and a circle

Solve the system $x^2 + y^2 = 25$ and $y = x + 1$ .

Substitute $x + 1$ for $y$ :

\begin{aligned} x^2 + (x + 1)^2 &= 25 \\ 2x^2 + 2x + 1 &= 25 \\ 2x^2 + 2x - 24 &= 0 \\ x^2 + x - 12 &= 0 \\ (x + 4)(x - 3) &= 0 \end{aligned}

So $x = 3$ or $x = -4$ . From $y = x + 1$ , the solutions are $(3, 4)$ and $(-4, -3)$ . Check: $9 + 16 = 25$ and $16 + 9 = 25$ .

The line y = x + 1 crosses the circle x² + y² = 25 at (3, 4) and (-4, -3).Open in grapher →

Remember to expand $(x + 1)^2$ fully as $x^2 + 2x + 1$ ; writing $x^2 + 1$ is a common slip.

Using the discriminant with a parameter

Because the number of solutions depends on the sign of the discriminant, you can find which lines are tangent without graphing.

Worked example: Finding a tangent line

For what value of $k$ is the line $y = 2x + k$ tangent to the parabola $y = x^2$ ?

Substitute: $x^2 = 2x + k$ , or $x^2 - 2x - k = 0$ . The line is tangent when there is exactly one solution, so the discriminant must be $0$ :

(-2)^2 - 4(1)(-k) = 4 + 4k = 0 \quad\Longrightarrow\quad k = -1.

Then $x^2 - 2x + 1 = (x - 1)^2 = 0$ , so the point of tangency is $(1, 1)$ .

Tip

If $k > -1$ in the last example, then $4 + 4k > 0$ and the line crosses the parabola twice; if $k < -1$ , the line misses it. Picture sliding the line $y = 2x + k$ up and down: it moves from missing, to touching, to cutting through.

Practice

Practice 1

Solve the system $y = x^2$ and $y = x + 6$ . Enter the x-coordinates of the intersection points.

Separate answers with commas, e.g. 2, -5

Practice 2

Solve the system $y = x^2 - 2x$ and $y = 2x - 4$ .

Enter a point like (2, -3)

Practice 3

How many solutions does the system $y = x^2 + 4$ and $y = -x + 1$ have?

Practice 4

The system $y = 2x^2 - 3x - 1$ and $y = x + 5$ has two solutions. Find the one with a positive x-coordinate.

Enter a point like (2, -3)

Practice 5

The system $x^2 + y^2 = 10$ and $y = x + 2$ has two solutions. Find the one in Quadrant III.

Enter a point like (2, -3)

Practice 6

A ball follows the path $y = -x^2 + 6x$ , where $x$ is horizontal distance and $y$ is height, both in meters. It lands on a ramp along the line $y = x$ . Apart from the starting point $(0, 0)$ , at what height does the ball hit the ramp?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

The line $y = kx$ with $k > 0$ is tangent to the parabola $y = x^2 + 1$ . Find $k$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

For which values of $b$ does the line $y = 4x + b$ intersect the parabola $y = x^2$ in two points? Write an inequality in $b$ .

Enter an inequality, e.g. x >= 4, -2 < x <= 3, or x < 1 or x > 5