Intersection of a quadratic function and a line

A point where the parabola $y = x^2$ meets a line is a point that sits on the parabola and on the line at the same time. At such a point the $y$ you get from the parabola and the $y$ you get from the line are exactly equal. So to find an intersection we set those two $y$ values equal and combine them into a single equation.

Solving for the intersections

Take the line $y = 2x + 3$ .

\begin{align*} x^2 &= 2x + 3 \\ x^2 - 2x - 3 &= 0 \\ (x - 3)(x + 1) &= 0 \end{align*}

So $x = 3$ or $x = -1$ . Putting these back into the parabola gives $y = 9$ and $y = 1$ , so the two intersection points are $(3, 9)$ and $(-1, 1)$ .

Counting without solving

The key idea is that the number of solutions of the equation is the number of intersection points. You can find that count using the discriminant $D = b^2 - 4ac$ of the quadratic $ax^2 + bx + c = 0$ ¹. It is the part under the square root in the quadratic formula, and its sign decides everything.

Discriminant	Solutions	The line and the parabola
$D > 0$	two	cut at two points
$D = 0$	one	are tangent
$D < 0$	none	do not meet

Sliding a line of fixed slope

If you slide a line of the same slope up and down, these three cases appear in turn: while the line runs deep through the parabola there are two crossings, as it drops the two crossings move closer, at one moment they merge into a single point where the line is tangent, and lower still the line leaves the parabola entirely.

Three lines of slope $2$ are drawn on the graph, each giving $x^2 - 2x - k = 0$ with discriminant $D = 4 + 4k$ .

Line	Equation	$D$	Result
$y = 2x + 3$	$x^2 - 2x - 3 = 0$	$16$	two crossings
$y = 2x - 1$	$(x - 1)^2 = 0$	$0$	tangent at $(1, 1)$
$y = 2x - 3$	$x^2 - 2x + 3 = 0$	$-8$	no crossing

The large dots on the graph mark the intersection and tangency points.

Discriminant, Wikipedia