Solve intersections of linear and quadratic functions

Ontario Grade 11 Mathematics

Characteristics of Functions

Finding Where a Line Meets a Parabola

Two functions intersect wherever their output values are equal for the same input. In this lesson you will find those shared points for one linear function and one quadratic function, a common and useful problem in Grade 11 mathematics. You already know how to solve quadratic equations by factoring and by the quadratic formula from Grade 10. Here you will use those skills in a new setting: combining two equations into one before solving. Work through each section in order, because every new idea builds on the one before it.

What you will learn

Understand what it means geometrically for a line and a parabola to intersect.
Set up and solve a system by substituting a linear function into a quadratic function.
Interpret the number of solutions using the discriminant.
Verify intersection points by substituting back into both original equations.

Prerequisite Bridge: What You Need to Remember

A linear function has the form

f(x) = mx + b

, where

m

is the slope and

b

is the y-intercept. Its graph is a straight line.

A quadratic function has the form

g(x) = ax^2 + bx + c

with

a \neq 0

. Its graph is a parabola, a smooth, symmetric curve that opens upward when

a > 0

and downward when

a < 0

.

To solve a quadratic equation

ax^2 + bx + c = 0

, you can factor it when the trinomial factors nicely, or you can use the quadratic formula shown below. The expression under the square root,

b^2 - 4ac

, is called the discriminant, and it tells you how many real solutions exist.

The discriminant rules are: if

b^2 - 4ac > 0

, there are two distinct real solutions; if

b^2 - 4ac = 0

, there is exactly one real solution, called a repeated root; if

b^2 - 4ac < 0

, there are no real solutions.

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Linear functions graph as straight lines; quadratic functions graph as parabolas.
The quadratic formula always works, even when factoring is difficult.
The discriminant $b^2 - 4ac$ predicts the number of intersection points before you fully solve.

What Does an Intersection Mean?

An intersection point of two functions is a point

(x, y)

that lies on both graphs at the same time. At that point, both functions produce exactly the same output for the same input value of

x

. In other words,

f(x) = g(x)

.

Geometrically, a straight line can cross a parabola in three distinct situations: it can miss the parabola entirely giving zero intersections, it can just touch the parabola at one point called a tangent point giving one intersection, or it can cut through the parabola at two separate points giving two intersections. Sketching a rough graph before you calculate is a good habit because it helps you predict which situation you are in.

The strategy for finding intersections is always the same: set the two function rules equal to each other, rearrange everything to one side so you have a quadratic equation equal to zero, and then solve that equation. Once you have the

x

-value or values, substitute back into the simpler linear function to find the matching

y

-value or values.

An intersection satisfies both functions simultaneously: $f(x) = g(x)$ .
A line and a parabola can have 0, 1, or 2 intersection points.
The method is set equal, rearrange to standard quadratic form, solve, then find $y$ -values.

The Substitution Method Step by Step

Step 1: set the functions equal. Write

f(x) = g(x)

, replacing each side with its rule. This is valid because you are looking for

x

-values where the outputs are identical.

Step 2: rearrange into standard form. Move all terms to one side so the equation looks like

ax^2 + bx + c = 0

. Be careful with signs when you move terms across the equals sign.

Step 3: choose a solution method. Check whether the trinomial factors neatly. If it does not factor, use the quadratic formula. You may also check the discriminant first to decide how many solutions to expect.

Step 4: solve for

x

. If there are two

x

-values, each one gives a separate intersection point. If the discriminant is negative, the line and parabola do not intersect.

Step 5: find

y

. Substitute each

x

-value into the linear function

f(x) = mx + b

to calculate the

y

-coordinate. Using the linear function is easier than the quadratic because it has fewer terms, though either function should give the same

y

-value, so you can use the quadratic to double-check.

Setting $f(x) = g(x)$ is the core of the substitution method.
Always rearrange to $ax^2 + bx + c = 0$ before applying a solving technique.
Substitute $x$ into the linear function to find $y$ ; it is faster and less error-prone.
Verification using both equations confirms your answer is correct.

Using the Discriminant to Predict Intersections

Once you have rearranged to

ax^2 + bx + c = 0

, compute

D = b^2 - 4ac

before solving. This is a quick check that saves time: if

D < 0

, stop, since there are no real intersections and further calculation is unnecessary.

If

D = 0

, the line is tangent to the parabola. It touches at exactly one point, and the quadratic has a single repeated root. If

D > 0

, you will find two distinct

x

-values and therefore two distinct intersection points.

This approach is especially useful on problems where the numbers are messy. Computing

D

first tells you whether to expect a clean answer or to trust an irrational, meaning a square-root, result.

D = b^2 - 4ac

$D = b^2 - 4ac$ predicts the number of intersections: $D < 0$ means none, $D = 0$ means one, $D > 0$ means two.
Checking the discriminant before solving is a useful efficiency strategy.
A tangent line produces a discriminant of exactly zero.

Connecting the Algebra to the Graph

After you solve algebraically, it is worth checking your answer visually. Plot the parabola and line roughly; even a sketch is enough. The

x

-values you calculated should correspond to the horizontal positions where the two graphs meet.

If your two

x

-values are, say,

x = -1

and

x = 3

, those are the two places on the

x

-axis directly below or above the intersection points. The

y

-values you calculated by substituting into the linear function tell you how high those intersection points sit.

Connecting algebra to a graph is not just a check; it builds your mathematical intuition. Over time, looking at a system of equations and picturing roughly where the graphs cross becomes a natural first step, making it easier to spot errors and understand solutions.

Always link algebraic solutions back to their geometric meaning on a graph.
Each $x$ -value solution corresponds to a horizontal crossing position on the graph.
A quick sketch before or after solving helps you catch sign errors and misread equations.

Number of Intersections: What the Discriminant Tells You

Discriminant $D = b^2 - 4ac$	Number of Intersections	Geometric Picture	What to Do Next
$D < 0$	0	Line misses the parabola entirely	Stop; no real solution exists
$D = 0$	1	Line is tangent to the parabola	One repeated $x$ -value; find one $(x, y)$ point
$D > 0$	2	Line cuts through the parabola	Two distinct $x$ -values; find two $(x, y)$ points

Worked example

Example 1: Two Intersection Points Using the Quadratic Formula

Find all points where

f(x) = 2x + 3

and

g(x) = x^2 - x - 3

intersect.

Set the functions equal
You want the $x$ -values where both functions give the same output, so write $f(x) = g(x)$ .
$2x + 3 = x^2 - x - 3$
Rearrange to standard quadratic form
Subtract $2x$ and $3$ from both sides so that the right side equals zero. This gives you a quadratic equation you can solve.
$0 = x^2 - 3x - 6$
Check the discriminant
Here $a = 1$ , $b = -3$ , $c = -6$ . Compute $D = (-3)^2 - 4(1)(-6) = 9 + 24 = 33$ . Since $D = 33 > 0$ , there are two distinct real intersections. However, $33$ is not a perfect square, so the trinomial does not factor over the integers, meaning the quadratic formula is needed.
$D = 9 + 24 = 33$
Apply the quadratic formula
Substitute $a = 1$ , $b = -3$ , $c = -6$ into the quadratic formula to find both $x$ -values.
$x = \frac{3 \pm \sqrt{33}}{2}$
Find the two x-values
The two solutions are $x_1 = \frac{3 + \sqrt{33}}{2}$ and $x_2 = \frac{3 - \sqrt{33}}{2}$ . Using a calculator, $\sqrt{33} \approx 5.745$ , so $x_1 \approx 4.37$ and $x_2 \approx -1.37$ .
$x_1 \approx 4.37, \ x_2 \approx -1.37$
Find the y-values using the linear function
Substitute each $x$ -value into $f(x) = 2x + 3$ because it is simpler to evaluate than the quadratic. For $x_1 \approx 4.37$ , $y_1 \approx 11.74$ . For $x_2 \approx -1.37$ , $y_2 \approx 0.26$ .
$y_1 \approx 11.74, \ y_2 \approx 0.26$

Answer: The two intersection points are approximately

(4.37, 11.74)

and

(-1.37, 0.26)

. In exact form:

\left(\frac{3+\sqrt{33}}{2}, 3+\sqrt{33}\right)

and

\left(\frac{3-\sqrt{33}}{2}, 3-\sqrt{33}\right)

.

Check: Substitute

x_1 \approx 4.37

into

g(x) = x^2 - x - 3

:

g(4.37) \approx 19.10 - 4.37 - 3 = 11.73

, which matches

y_1 \approx 11.74

within rounding. The same check works for

x_2

.

Worked example

Example 2: Exactly Two Intersection Points by Factoring

The quadratic function is

g(x) = x^2 - 4x + 7

and the linear function is

f(x) = 2x - 1

. Determine whether the line intersects the parabola and, if so, find the intersection point or points.

Set the functions equal
Write $f(x) = g(x)$ to find the $x$ -value or values where both functions share the same output.
$2x - 1 = x^2 - 4x + 7$
Rearrange to standard form
Subtract $2x$ and add $1$ to both sides to collect all terms on one side, so the equation equals zero.
$0 = x^2 - 6x + 8$
Check the discriminant
Identify $a = 1$ , $b = -6$ , $c = 8$ . Compute $D = (-6)^2 - 4(1)(8) = 36 - 32 = 4$ . Since $D = 4 > 0$ , there are two distinct intersection points, and because $D$ is a perfect square, the quadratic should factor neatly.
$D = 36 - 32 = 4$
Factor the quadratic
Look for two numbers that multiply to $8$ and add to $-6$ . Those numbers are $-2$ and $-4$ , so the equation factors as shown.
$(x - 2)(x - 4) = 0$
Solve for x
Setting each factor equal to zero gives the two $x$ -values where the line and parabola meet: $x = 2$ or $x = 4$ .
$x - 2 = 0, \quad x - 4 = 0$
Find the y-values using the linear function
Substitute each $x$ -value into $f(x) = 2x - 1$ . For $x = 2$ , $y = 2(2) - 1 = 3$ . For $x = 4$ , $y = 2(4) - 1 = 7$ .
$y = 2x - 1$

Answer: The line intersects the parabola at two points:

(2, 3)

and

(4, 7)

.

Check: Check

(2, 3)

in

g(x)

:

g(2) = 4 - 8 + 7 = 3

, which matches. Check

(4, 7)

in

g(x)

:

g(4) = 16 - 16 + 7 = 7

, which matches. Both points satisfy the quadratic function, confirming the solution.

Common mistakes and how to avoid them

Forgetting to move all terms to one side before solving, leaving something like

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

and trying to factor it as written.

Correction: Always rearrange to

ax^2 + bx + c = 0

first. Every factoring and formula technique requires zero on one side.

Substituting the

x

-value back into the wrong equation and getting a different

y

, then thinking the answer is wrong.

Correction: Both functions must give the same

y

at an intersection. If they do not, an arithmetic error was made somewhere. Use both functions as a cross-check.

Making sign errors when rearranging, such as moving

-x

to the left and writing

+x

instead, or mishandling a negative constant.

Correction: Write out every term on its own line when rearranging. Reverse the sign of every term you move across the equals sign.

Identifying

a

,

b

,

c

incorrectly for the discriminant or formula, especially using the original quadratic's coefficients rather than those of the rearranged equation.

Correction: Read

a

,

b

,

c

from the rearranged equation

ax^2 + bx + c = 0

, not from the original separate functions.

Stopping after finding the

x

-values and forgetting to calculate the

y

-values, leaving the answer incomplete.

Correction: An intersection point is a coordinate pair

(x, y)

. Always substitute each

x

into the linear function to find the corresponding

y

.

Lesson summary

To find intersections of a linear function $f(x)$ and a quadratic function $g(x)$ , set $f(x) = g(x)$ and rearrange to standard quadratic form $ax^2 + bx + c = 0$ .
The discriminant $D = b^2 - 4ac$ tells you the number of intersections: negative means none, zero means one and tangent, positive means two.
Solve the resulting quadratic by factoring when possible, or by the quadratic formula when the trinomial does not factor neatly.
Substitute each $x$ -solution into the simpler linear function to find the $y$ -coordinate of each intersection point.
Always verify your answer by checking that both original functions produce the same $y$ -value at each intersection point.
Write each final answer as an ordered pair $(x, y)$ and connect it to the geometric picture, the point where the line and parabola meet on the graph.

Check your understanding

Question 1

What is the first algebraic step when finding the intersection of

f(x) = 3x + 5

and

g(x) = x^2 + x - 2

?

Factor $x^2 + x - 2$ immediately.
Set $3x + 5 = x^2 + x - 2$ and rearrange to equal zero.
Graph both functions on the same axes.
Substitute $x = 0$ into both functions.

Show answer and explanation

Set

3x + 5 = x^2 + x - 2

and rearrange to equal zero.

The substitution method begins by setting the two function rules equal to each other. You then rearrange so that all terms are on one side, giving a quadratic equation equal to zero, which you can then solve.

Question 2

After rearranging a system to

x^2 - 6x + 9 = 0

, what does the discriminant tell you?

There are two distinct intersection points.
There are no real intersection points.
The line is tangent to the parabola at exactly one point.
The parabola opens downward.

Show answer and explanation

The line is tangent to the parabola at exactly one point.

D = (-6)^2 - 4(1)(9) = 36 - 36 = 0

. A discriminant of zero means there is exactly one repeated solution, so the line just touches the parabola at one point, meaning it is tangent to it.

Question 3

You find

x = -2

and

x = 5

as the solutions. The linear function is

f(x) = x + 4

. What are the two intersection points?

$(-2, -2)$ and $(5, 5)$
$(-2, 2)$ and $(5, 9)$
$(-2, 6)$ and $(5, 1)$
$(-2, -6)$ and $(5, 9)$

Show answer and explanation

(-2, 2)

and

(5, 9)

Substitute each

x

-value into

f(x) = x + 4

. For

x = -2

,

f(-2) = 2

, giving

(-2, 2)

. For

x = 5

,

f(5) = 9

, giving

(5, 9)

.

Question 4

A student rearranges a system and gets

2x^2 - 3x + 8 = 0

. The discriminant is

D = 9 - 64 = -55

. What is the correct conclusion?

There are two intersection points with irrational coordinates.
The student made an error; every quadratic has at least one solution.
The line and parabola do not intersect; there are no real solutions.
The line is tangent to the parabola.

Show answer and explanation

The line and parabola do not intersect; there are no real solutions.

A negative discriminant,

D = -55 < 0

, means the quadratic equation has no real solutions. Geometrically, this means the line misses the parabola entirely; they share no common points.

Key terms

Intersection point: A point $(x, y)$ that lies on both graphs simultaneously; both functions produce the same output $y$ for the same input $x$ .
Linear function: A function of the form $f(x) = mx + b$ whose graph is a straight line with slope $m$ and y-intercept $b$ .
Quadratic function: A function of the form $g(x) = ax^2 + bx + c$ with $a \neq 0$ whose graph is a parabola.
Standard quadratic form: The arrangement $ax^2 + bx + c = 0$ , with all terms on one side and zero on the other, required before applying factoring or the quadratic formula.
Discriminant: The expression $D = b^2 - 4ac$ inside the quadratic formula. Its sign tells you whether a quadratic equation has two, one, or no real solutions.
Tangent point: The single point where a line just touches a curve without crossing through it, corresponding to a discriminant of exactly zero.
Substitution method: A technique for solving a system of equations by replacing one equation's expression into the other, reducing the system to a single equation in one variable.
Quadratic formula: The formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ used to find the solutions of any quadratic equation in standard form.

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