Intersections of Quadratic Graphs

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

1 point of intersection

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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