A. Characteristics of Functions · Quadratic Functions

Where a Line Meets a Parabola

A2.5About 70 minutes

Learning goals

I can find the intersection points of a linear function and a quadratic function algebraically.
I can find and check them graphically.
I can use the discriminant to tell whether a line crosses, touches or misses a parabola.
I can solve a real-world problem that involves a line and a parabola.

Words to know

linear-quadratic system
A linear function and a quadratic function considered together; its solutions are their intersection points.

point of intersection
A point that lies on both graphs.

secant line
A line that crosses a curve at two points.

tangent line
A line that touches a parabola at exactly one point without crossing it.

Where paths cross

A snowball arcs through the air toward a snowy hillside. Where does it hit the slope? The ball follows a parabola; the hillside is close to a straight line. The landing point is where the two graphs intersect.

The curriculum asks you to solve problems about the intersection of a linear function and a quadratic function, both graphically and algebraically. You already know how to solve linear systems (two lines) from Grade 10. A linear-quadratic system works the same way, but it can have two, one or no solutions.

Solving algebraically

At an intersection point, both functions have the same output, so f(x) = g(x).

Set the two expressions equal.
Move everything to one side to get a quadratic equation, ax² + bx + c = 0.
Solve it (factor, or use the quadratic formula).
Substitute each x into the linear function (it is easier) to find y.
Check each point in the quadratic function.

Worked example

Example 1: Two intersection points

Find the intersection points of f(x) = x² − 4x + 5 and g(x) = x + 1.

Set them equal: x² − 4x + 5 = x + 1.
Rearrange: x² − 5x + 4 = 0.
Factor: (x − 1)(x − 4) = 0, so x = 1 or x = 4.
Find y with g: g(1) = 2 and g(4) = 5.
Check in f: f(1) = 1 − 4 + 5 = 2 ✓ and f(4) = 16 − 16 + 5 = 5 ✓.

The graphs meet at (1, 2) and (4, 5).

Solving graphically: the line g(x) = x + 1 (dashed) crosses the parabola f(x) = x² − 4x + 5 at (1, 2) and (4, 5), the same points found algebraically.

How many intersections?

The quadratic equation from step 2 decides it, through its discriminant:

D > 0: the line crosses the parabola at two points (a secant line).
D = 0: the line touches the parabola at exactly one point (a tangent line).
D < 0: the line misses the parabola.

(A vertical line, like x = 3, meets a parabola that opens up or down exactly once, but a vertical line is not a linear function.)

Three ways a line and a parabola can meet. Left: y = x + 1 and y = x² − 4x + 5. Middle: y = 2x − 1 is tangent to y = x² at (1, 1). Right: y = x never meets y = x² + 3.

Worked example

Example 2: One point

Solve the system f(x) = x², g(x) = 2x − 1.

x² = 2x − 1, so x² − 2x + 1 = 0.
(x − 1)² = 0, so x = 1 (D = 4 − 4 = 0).
g(1) = 1.

One point, (1, 1). The line is tangent to the parabola there.

Worked example

Example 3: No points

Do f(x) = x² + 3 and g(x) = x intersect?

x² + 3 = x, so x² − x + 3 = 0.
D = (−1)² − 4(1)(3) = 1 − 12 = −11 < 0.

No. The line passes below the parabola everywhere.

Worked example

Example 4: A parabola opening down

Solve f(x) = −x² + 2x + 8 and g(x) = 2x − 1.

−x² + 2x + 8 = 2x − 1.
The 2x terms cancel: −x² + 9 = 0, so x² = 9 and x = ±3.
g(3) = 5 and g(−3) = −7.
Check: f(3) = −9 + 6 + 8 = 5 ✓ and f(−3) = −9 − 6 + 8 = −7 ✓.

(3, 5) and (−3, −7).

Worked example

Example 5: A snowball and a hill

A snowball is thrown from 1 m above the base of a hill. Its height is h(x) = −0.05x² + x + 1 metres, where x is the horizontal distance in metres. The hillside rises along the line y = 0.2x. Where does the snowball hit the hill?

Set the heights equal: −0.05x² + x + 1 = 0.2x.
−0.05x² + 0.8x + 1 = 0. Multiply by −20: x² − 16x − 20 = 0.
x = (16 ± √(256 + 80))/2 = (16 ± √336)/2. √336 = √(16 × 21) = 4√21, so x = 8 ± 2√21.
Only the positive value makes sense: x = 8 + 2√21 ≈ 17.17.
Height: y = 0.2(17.17) ≈ 3.43.

The snowball hits the hill about 17.2 m away horizontally, at a height of about 3.4 m.

The snowball's path (solid) and the hillside (dashed). The intersection is the landing point, about (17.2, 3.4).

Graph or algebra?

A graph (by hand or with graphing technology) shows how many solutions there are and roughly where. Algebra gives exact answers, like x = 8 + 2√21. A good habit: estimate from a sketch, then solve algebraically and check that the answers agree.

Hands-on activity

Ruler and parabola

You need: graph paper, a pencil and a clear ruler (or a straight strip of cardboard).

Plot y = x² carefully for x = −3 to 3 on graph paper (1 square = 1 unit).
Lay the ruler along the line y = x + 2 (through (0, 2) and (2, 4)). Mark where it crosses the parabola and read the coordinates.
Solve x² = x + 2 algebraically. Do your readings match?
Keep the ruler at the same slope (parallel) and slide it down until it just touches the parabola. Read the point where it touches.
Slide it a little lower. How many intersection points now?

What to look for: y = x + 2 crosses at (−1, 1) and (2, 4), since x² − x − 2 = (x − 2)(x + 1). Lines of slope 1 are y = x + b; x² − x − b = 0 has D = 1 + 4b, so the touching line is b = −1/4, touching at (0.5, 0.25). Any lower and there are no intersections.

Just for fun

A line met a parabola at a party. "Let's keep this to one point of contact," said the line. "Ah," said the parabola, "so you're going to be tangent-ial."

Your turn

Find the intersection points of f(x) = x² − 2x − 3 and g(x) = x + 1.

Solve the system f(x) = x² + 1, g(x) = 2x.

Do f(x) = −x² + 4 and g(x) = x + 5 intersect?

Find the intersection points of f(x) = 2x² − 3x − 1 and g(x) = x + 5.

Find the intersection points of f(x) = x² − 4x and g(x) = −2x + 3.

Find the intersection points of f(x) = x² + 2x − 1 and g(x) = 3x + 1.

More practice: warm-up

What is the greatest number of points where a line (a linear function) can meet a parabola?

Is (2, 4) on both y = x² and y = 2x?

Solve x² = x + 6.

Find the intersection points of y = x² and y = x + 6.

After setting f(x) = g(x) and rearranging, you get D < 0. What does that mean?

More practice: on your own

Find the exact intersection points of f(x) = x² − 3 and g(x) = x.

Find the intersection points of f(x) = −2x² + 5x + 3 and g(x) = −x + 3.

Where does the horizontal line y = 4 meet f(x) = x² − 6x + 9?

A snowball's height is h(x) = −0.1x² + 1.2x + 1.6 metres, thrown toward a slope y = 0.4x. Where does it land? Give the exact x, then round.

A company's revenue is R(x) = −2x² + 60x and its cost is C(x) = 10x + 200, both in thousands of dollars, for x thousand items. Find the break-even points.

Challenge

The line y = x + 1 cuts the parabola f(x) = x² − 4x + 5 at two points (Example 1). How long is the segment between them? Give an exact answer.

Show that the line y = 4x − 6 is tangent to y = x² − 2x + 3, and find the point of tangency.

A line with slope 3 passes through the vertex of y = (x − 2)² + 1. Find the other point where it meets the parabola.

Check your understanding

Quiz

Where do y = x² and y = x + 2 intersect?

Which number of intersection points is impossible for a non-vertical line and a parabola?

After finding x-values of intersection points, the easiest way to find the y-values is usually to:

How many times does g(x) = 2x meet f(x) = x² + 5?

The line g(x) = 6x − 9 and the parabola f(x) = x² meet at:

Which equation gives the x-coordinates where f(x) = 2x² + x meets g(x) = 3x + 4?

For parents and teachers

Big idea: Expectation A2.5: solve problems involving the intersection of a linear function and a quadratic function, graphically and algebraically.

How to help: Ask your teen to sketch both graphs before solving, predict how many intersection points there should be, and check each answer in both equations.

Common difficulties:

Forgetting to move every term to one side before factoring.
Finding x but not y (a point has two coordinates).
Sign errors when subtracting the linear expression, such as x² − 4x + 5 − (x + 1).
Keeping a solution that makes no sense in context, such as a negative distance.