How many times does the line $y = x + 2$ meet the parabola $y = x^2 - 3x + 6$?

Once: the line is tangent to the parabola at $(2, 4)$.

Solve $y = 2x^2 + 3x - 4$ and $y = x + 1$, giving exact values.

$\left(\frac{-1 + \sqrt{11}}{2}, \frac{1 + \sqrt{11}}{2}\right) \approx (1.16, 2.16)$ and $\left(\frac{-1 - \sqrt{11}}{2}, \frac{1 - \sqrt{11}}{2}\right) \approx (-2.16, -1.16)$.

Without solving, decide whether the line $3x - y = -1$ meets the parabola $y = -x^2 + 2x$.

No: the discriminant is negative, so the line misses the parabola.

Find $k$ so that the line $y = 2x + k$ is tangent to $y = x^2 - 4x + 1$, and find the point of tangency. For which $k$ does the line miss the parabola?

$k = -8$, tangent at $(3, -2)$; the line misses the parabola when $k < -8$.

Grade 11 · MCR3U · Functions

MCR3U: Linear-Quadratic Systems

A linear-quadratic system asks where a line meets a parabola. A line can cross a parabola twice, touch it once or miss it completely, so the system has two, one or no solutions. Substitution finds the points, and the discriminant counts them.

How It Works

Substitute. With $y = x^2 - 4x + 3$ and $y = 2x - 5$ , set the right sides equal: $x^2 - 4x + 3 = 2x - 5$ . If the line comes as $2x - y = 5$ , solve it for $y$ first.

Move everything to one side. $x^2 - 6x + 8 = 0$ is an ordinary quadratic equation. Factor it, or use the quadratic formula when it does not factor.

Find both coordinates. $(x - 2)(x - 4) = 0$ gives $x = 2$ and $x = 4$ . Substitute each into the linear equation, which is the easier one: $y = -1$ and $y = 3$ . The points of intersection are $(2, -1)$ and $(4, 3)$ .

The discriminant counts the intersections. For the combined equation $ax^2 + bx + c = 0$ , work out $D = b^2 - 4ac$ . If $D > 0$ the line is a secant and crosses twice. If $D = 0$ it is a tangent and touches once. If $D < 0$ the two never meet. You can say how many without solving.

Unknown constants. A common test question gives the line as $y = mx + k$ with one letter unknown and asks for the value that makes it tangent. Substitute, collect terms, and set $D = 0$ : the discriminant becomes an equation in that letter. Use $D > 0$ or $D < 0$ for two points or none.

Check each point in both equations. Substitute into the quadratic as well. It catches sign slips that the linear equation hides.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
Solve the system $y = x^2 - 1$ and $y = x + 1$ .
Show solution
1. Set equal: $x^2 - 1 = x + 1$ , so $x^2 - x - 2 = 0$ .
2. Factor: $(x - 2)(x + 1) = 0$ , so $x = 2$ or $x = -1$ .
3. From $y = x + 1$ : $y = 3$ and $y = 0$ .
Answer: $(-1, 0)$ and $(2, 3)$ .
Easy
How many times does the line $y = x + 2$ meet the parabola $y = x^2 - 3x + 6$ ?
Show solution
1. Set equal and collect: $x^2 - 4x + 4 = 0$ .
2. $D = (-4)^2 - 4(1)(4) = 0$ , so the line is tangent.
3. $(x - 2)^2 = 0$ gives $x = 2$ and $y = 4$ .
Answer: Once: the line is tangent to the parabola at $(2, 4)$ .
Medium
Solve $y = 2x^2 + 3x - 4$ and $y = x + 1$ , giving exact values.
Show solution
1. Set equal and collect: $2x^2 + 2x - 5 = 0$ , which does not factor.
2. Quadratic formula: $x = \frac{-2 \pm \sqrt{4 + 40}}{4} = \frac{-2 \pm 2\sqrt{11}}{4} = \frac{-1 \pm \sqrt{11}}{2}$ .
3. $y = x + 1 = \frac{1 \pm \sqrt{11}}{2}$ , taking the same sign.
Answer: $\left(\frac{-1 + \sqrt{11}}{2}, \frac{1 + \sqrt{11}}{2}\right) \approx (1.16, 2.16)$ and $\left(\frac{-1 - \sqrt{11}}{2}, \frac{1 - \sqrt{11}}{2}\right) \approx (-2.16, -1.16)$ .
Medium
Without solving, decide whether the line $3x - y = -1$ meets the parabola $y = -x^2 + 2x$ .
Show solution
1. Solve the line for $y$ : $y = 3x + 1$ .
2. Set equal: $-x^2 + 2x = 3x + 1$ , so $x^2 + x + 1 = 0$ .
3. $D = 1^2 - 4(1)(1) = -3 < 0$ .
Answer: No: the discriminant is negative, so the line misses the parabola.
Test level
Find $k$ so that the line $y = 2x + k$ is tangent to $y = x^2 - 4x + 1$ , and find the point of tangency. For which $k$ does the line miss the parabola?
Show solution
1. Set equal and collect: $x^2 - 6x + (1 - k) = 0$ .
2. $D = (-6)^2 - 4(1)(1 - k) = 32 + 4k$ .
3. Tangent when $D = 0$ : $k = -8$ . Then $x^2 - 6x + 9 = 0$ , so $x = 3$ and $y = 2(3) - 8 = -2$ .
4. Check in the parabola: $9 - 12 + 1 = -2$ .
5. The line misses when $D < 0$ : $32 + 4k < 0$ , so $k < -8$ .
Answer: $k = -8$ , tangent at $(3, -2)$ ; the line misses the parabola when $k < -8$ .

Learn It on Martian Lab

Martian Lab teaches this in the Linear-Quadratic Systems lesson of its MCR3U Quadratic Functions unit. Watch it, then take the Quadratic Functions practice tests to see if it stuck.

Watch the Linear-Quadratic Systems lesson free

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Free Quadratic Functions practice test, with solutions

More Quadratic Functions topics: Max/Min and Vertex Form · Quadratic Word Problems · Simplifying Radicals

More MCR3U topics: Inverse Functions · Transformations of Functions · Sine Law and the Ambiguous Case · Domain, Range and Function Notation · Simplifying Rational Expressions · Operations with Rational Expressions · Exponent Laws and Rational Exponents · Graphing Exponential Functions · Exponential Growth and Decay · Special Angles and the CAST Rule · Cosine Law and 3D Problems · Transformations of Sine and Cosine · Sinusoidal Word Problems · Arithmetic Sequences and Series · Geometric Sequences and Series · Pascal's Triangle and the Binomial Theorem · Simple and Compound Interest · Annuities: Future and Present Value

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