Before you read

Answer first — see what you already know.

A laser beam at height $y = 4$ intersects a parabolic mirror $y = x^{2}$ . How many intersection points are there?

A
$0$
B
$1$
C
$2$
D
$4$

Check your answer

Answer:

Before you read

Answer first — see what you already know.

Find the intersection points of a parabolic arch $y = x^{2}$ and a ramp $y = 3 x - 2$ .

A
$(1, 1)$ and $(2, 4)$
B
$(1, 1)$ only
C
$(- 1, 1)$ and $(2, 4)$
D
No solution

Check your answer

Answer:

Theory

Quadratic-Linear Systems

A quadratic-linear system pairs a parabola with a line:
$\begin{cases} y = x^2 + bx + c \\ y = mx + d \end{cases}$

Graphically, you're finding where the line intersects the parabola.

y = x² intersecting with a line

There can be:
- 2 intersections (line cuts through parabola)
- 1 intersection (line is tangent to parabola)
- 0 intersections (line misses parabola)

Solving: substitute the linear equation into the quadratic (or set them equal).

Theory

The Method

Since both equations equal $y$ , set them equal:

$x^2 + bx + c = mx + d$

Rearrange to standard form and solve the resulting quadratic.

Example 1

Solve: $y = x^2$ and $y = x + 2$

Set equal: $x^2 = x + 2$

Rearrange: $x^2 - x - 2 = 0$

Factor: $(x-2)(x+1) = 0$ $\to$ $x = 2$ or $x = -1$

Find $y$ : $y = 2 + 2 = 4$ and $y = -1 + 2 = 1$

Solutions: $(2, 4)$ and $(-1, 1)$ .

(2, 4)

and

(-1, 1)

Example 2

Solve: $y = x^2 - 4x + 5$ and $y = x - 1$

Set equal: $x^2 - 4x + 5 = x - 1$

Rearrange: $x^2 - 5x + 6 = 0$

Factor: $(x-2)(x-3) = 0$ $\to$ $x = 2$ or $x = 3$

$y = 2 - 1 = 1$ and $y = 3 - 1 = 2$

Solutions: $(2, 1)$ and $(3, 2)$ .

(2, 1)

and

(3, 2)

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