# Intersections of Quadratic Graphs

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

**cutting**

2 points of intersection

**touching**

1 point of intersection

**missing**

no points of intersection

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} \)

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