Linear–Quadratic Systems

A linear–quadratic system asks where a line and a parabola meet. It might be where a thrown ball lands on a sloped hill, or where a road crosses a curved path. You’ll solve these with algebra, and the discriminant tells you in advance how many meeting points there are.

Key ideas

Three possibilities

A line and a parabola can meet at two points, one point, or none. When they meet at exactly one point, the line just touches the parabola: it’s a tangent line.

The same parabola with three different lines.

Solving by substitution

1. Set the two expressions for $y$ equal to each other.
2. Rearrange into the form $ax^2 + bx + c = 0$.
3. Solve for $x$ (factoring or the quadratic formula).
4. Substitute each $x$ into the linear equation (it’s easier) to find $y$.
5. Write the answers as points $(x, y)$.

Counting intersections

The quadratic from step 2 decides everything. Its discriminant $D = b^2 - 4ac$ gives:

• $D \gt 0$: two intersection points
• $D = 0$: one point (the line is tangent)
• $D \lt 0$: no intersection points

Worked examples

Example 1: Two intersection points

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

Solution. Set the expressions equal and rearrange:

$$x^2 - 2 = x \quad\Rightarrow\quad x^2 - x - 2 = 0 \quad\Rightarrow\quad (x - 2)(x + 1) = 0$$

So $x = 2$ or $x = -1$. Using $y = x$, the points are $(2, 2)$ and $(-1, -1)$.

Check $(2, 2)$ in the parabola: $2^2 - 2 = 2$. ✓

Example 2: A tangent line

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

Solution.

$$x^2 - 2 = 2x - 3 \quad\Rightarrow\quad x^2 - 2x + 1 = 0 \quad\Rightarrow\quad (x - 1)^2 = 0$$

So $x = 1$ only, and $y = 2(1) - 3 = -1$. The line touches the parabola at the single point $(1, -1)$: it’s a tangent.

Example 3: Counting without solving

How many times does the line $y = x - 4$ meet the parabola $y = 2x^2 + 3x - 1$?

Solution.

$$2x^2 + 3x - 1 = x - 4 \quad\Rightarrow\quad 2x^2 + 2x + 3 = 0$$

$D = 2^2 - 4(2)(3) = 4 - 24 = -20 \lt 0$, so they never meet.

Example 4: Finding a tangent line

For which values of $k$ is the line $y = kx - 3$ tangent to the parabola $y = x^2 + 1$?

Solution.

$$x^2 + 1 = kx - 3 \quad\Rightarrow\quad x^2 - kx + 4 = 0$$

Tangent means exactly one point, so $D = 0$:

$$(-k)^2 - 4(1)(4) = 0 \quad\Rightarrow\quad k^2 = 16 \quad\Rightarrow\quad k = 4 \text{ or } k = -4$$

Common mistakes

Stopping at the $x$-values. The answer to a system is a set of points. Find each matching $y$.

Not rearranging to $= 0$. You can’t factor $x^2 - 2 = x$ as it stands. Move everything to one side first.

Taking the discriminant of the wrong equation. Use the combined equation from step 2, not the original parabola.

Sign errors when moving terms. Subtracting $kx - 3$ from both sides gives $-kx + 3$. Write every step.

Substituting into the wrong equation without checking. Either equation works for finding $y$, but checking the point in both catches mistakes.

Practice

1. (Warm-up) Is $(3, 7)$ a solution of the system $y = x^2 - 2$ and $y = 2x + 1$?

Solution

Parabola: $3^2 - 2 = 7$. ✓ Line: $2(3) + 1 = 7$. ✓

Yes, $(3, 7)$ is on both graphs.

2. (Warm-up) What is the greatest number of points where a line and a parabola can meet? Explain.

Solution

Two. Setting the equations equal gives a quadratic equation, which has at most two solutions.

3. (Warm-up) Solve the system $y = x^2$ and $y = 4$.

Solution

$x^2 = 4$, so $x = 2$ or $x = -2$. The points are $(2, 4)$ and $(-2, 4)$.

4. (Core) Solve the system $y = x^2 + 2x - 3$ and $y = x + 3$.

Solution

$$x^2 + 2x - 3 = x + 3 \quad\Rightarrow\quad x^2 + x - 6 = 0 \quad\Rightarrow\quad (x + 3)(x - 2) = 0$$

$x = -3$ gives $y = 0$, and $x = 2$ gives $y = 5$. The points are $(-3, 0)$ and $(2, 5)$.

5. (Core) Solve the system $y = -x^2 + 4$ and $y = -2x + 5$. What does your answer tell you about the line?

Solution

$$-x^2 + 4 = -2x + 5 \quad\Rightarrow\quad x^2 - 2x + 1 = 0 \quad\Rightarrow\quad (x - 1)^2 = 0$$

$x = 1$ and $y = -2(1) + 5 = 3$. There’s only one point, $(1, 3)$, so the line is tangent to the parabola.

6. (Core) How many points of intersection do $y = 3x^2 - x + 2$ and $y = 5x - 1$ have?

Solution

$$3x^2 - x + 2 = 5x - 1 \quad\Rightarrow\quad 3x^2 - 6x + 3 = 0$$

$D = (-6)^2 - 4(3)(3) = 36 - 36 = 0$, so exactly one point. (The line is tangent.)

7. (Core) A ball is thrown from the bottom of a hill. Its path is $y = -0.1x^2 + 1.2x + 2$, and the hill’s surface is $y = 0.4x$, where $x$ is the horizontal distance and $y$ the height, both in metres. Where does the ball land on the hill?

Solution

$$\begin{aligned} -0.1x^2 + 1.2x + 2 &= 0.4x \\ -0.1x^2 + 0.8x + 2 &= 0 \\ x^2 - 8x - 20 &= 0 && \text{multiply by } -10 \\ (x - 10)(x + 2) &= 0 \end{aligned}$$

The horizontal distance must be positive, so $x = 10$, and $y = 0.4(10) = 4$. The ball lands $10$ m out horizontally, at a height of $4$ m on the hill.

8. (Challenge) For which values of $k$ does the line $y = 2x + k$ miss the parabola $y = x^2 - 4x + 5$ completely?

Solution

$$x^2 - 4x + 5 = 2x + k \quad\Rightarrow\quad x^2 - 6x + (5 - k) = 0$$

No intersection means $D \lt 0$:

$$36 - 4(5 - k) \lt 0 \quad\Rightarrow\quad 16 + 4k \lt 0 \quad\Rightarrow\quad k \lt -4$$

9. (Challenge) Find the equation of the line with slope $3$ that is tangent to $y = x^2 + x$, and the point where it touches.

Solution

Let the line be $y = 3x + b$.

$$x^2 + x = 3x + b \quad\Rightarrow\quad x^2 - 2x - b = 0$$

Tangent means $D = 0$: $(-2)^2 - 4(1)(-b) = 4 + 4b = 0$, so $b = -1$.

The line is $y = 3x - 1$. Then $x^2 - 2x + 1 = 0$ gives $x = 1$, and $y = 3(1) - 1 = 2$, so it touches at $(1, 2)$.