Consider the graphs of a quadratic function and a linear function on the same set of axes.

cutting
2 points of intersection

$b^2-4ac \gt 0$

touching
1 point of intersection

$b^2-4ac = 0$

missing
no points of intersection

$b^2-4ac \lt 0$

In the graphs meet, the coordinates of the points of intersection of the graphs can be found by solving the two equations simultaneously.

### Example 1

Find the coordinates of the points of intersection of the graphs with equations $y=x^2$ and $y=x+2$.

\begin{align} \displaystyle x^2 &= x+2 \\ x^2 -x-2 &= 0 \\ (x-2)(x+1) &= 0 \\ x &= 2 \text{ or } x = -1 \\ x &= 2 \text{ giving } y = 2^2 = 4 \\ x &= -1 \text{ giving } y = (-1)^2 = 1 \\ \therefore &(2,4) \text{ and } (-1,1) \end{align}

### Example 2

Find the coordinates of the points of intersection of the graphs with equations $y=2x^2+2x-5$ and $y=x^2-2x$.

\begin{align} \displaystyle 2x^2+2x-5 &= y=x^2-2x \\ x^2 + 4x – 5 &= 0 \\ (x+5)(x-1) &= 0 \\ x &= -5 \text{ or } x = 1 \\ x &= -5 \text{ giving } y = (-5)^2 – 2 \times (-5) = 35 \\ x &= 1 \text{ giving } y = 1^2 – 2 \times 1 = -1 \\ \therefore &(-5,35) \text{ and } (1,-1) \end{align}

### Example 3

$y=x^2-x-2$ and $y=x-k$ have two intersections. Find $k$.

\begin{align} \displaystyle x^2-x-2 &= x-k \\ x^2 -2x+k-2 &= 0 \\ b^2-4ac &\gt 0 &\text{two intersections}\\ (-2)^2-4 \times 1 \times (k-2) > 0 \\ 4-4k+8 > 0 \\ -4k &\gt -12 \\ \therefore k &\lt 3 \end{align}