1. Write both equations in the form \(y = \dots\) or \(x = \dots\).
2. Substitute the linear equation into the quadratic.
3. Solve the resulting quadratic equation.
4. Substitute back to find the corresponding values of the other variable.

Find the points where

\[ y = x + 1 \quad \text{and} \quad y = x^2 - 4 \]

Step 1: Substitute \(y = x + 1\) into \(y = x^2 - 4\):

\[ x + 1 = x^2 - 4 \Rightarrow x^2 - x - 5 = 0 \]

Step 2: Solve the quadratic:

\[ x = \frac{1 \pm \sqrt{1 + 20}}{2} = \frac{1 \pm \sqrt{21}}{2} \]

Step 3: Substitute into \(y = x + 1\):

\[ y = \frac{1 \pm \sqrt{21}}{2} + 1 = \frac{3 \pm \sqrt{21}}{2} \]

How many intersection points do the curves have?

\[ y = 2x + 3, \quad y = x^2 - x + 1 \]

Substitute into the quadratic:

\[ 2x + 3 = x^2 - x + 1 \Rightarrow x^2 - 3x - 2 = 0 \]

Use the discriminant \( \Delta = b^2 - 4ac \):

\[ \Delta = (-3)^2 - 4(1)(-2) = 9 + 8 = 17 > 0 \]

Determine whether the line meets the curve:

\[ y = 4x + 1, \quad y = x^2 - 2x + 5 \]

Substitute:

\[ 4x + 1 = x^2 - 2x + 5 \Rightarrow x^2 - 6x + 4 = 0 \]

Discriminant:

\[ \Delta = (-6)^2 - 4(1)(4) = 36 - 16 = 20 > 0 \]

Does the line touch the curve?

\[ y = 3x + 4, \quad y = x^2 + 6x + 11 \]

Substitute:

\[ 3x + 4 = x^2 + 6x + 11 \Rightarrow x^2 + 3x + 7 = 0 \]

Discriminant:

\[ \Delta = 3^2 - 4(1)(7) = 9 - 28 = -19 < 0 \]

• Always substitute into the quadratic, not the line.
• Use the discriminant to check how many intersections:

• Solutions must be written in coordinate form \((x,y)\).

1. \(y = 5x - 1,\quad y = x^2 + 3x - 2\)
2. \(y = 2x + 6,\quad y = x^2 + 4x + 9\)
3. \(x + y = 3,\quad y = x^2 - x + 1\)

Find the intersection points or show that there are none.