To find where the line and parabola intersect, set their $y$-values equal and rearrange the resulting equation into standard quadratic form in $x$.

For the quadratic equation you get, recall what must be true about the discriminant $b^2 - 4ac$ if the equation has exactly one real solution.

Solving the discriminant equation will give more than one possible value of $m$. Make sure to choose the one that satisfies $m > 0$.

In Desmos, enter the parabola: y = x^2 - 6x + 13.

Type y = m x + 1 and when Desmos prompts you, create a slider for m.

Move the slider for m and watch how the line intersects the parabola. You are looking for the value of m where the line just touches the parabola at exactly one point (the two graphs meet but do not cross).

Once you find the position where the line is tangent to the parabola, look at the slider to see the corresponding value of m. That is the slope that makes the system have exactly one solution.

The intersection points of the line and the parabola occur where their $y$-values are equal.

Set

Rearrange everything to one side to make a quadratic in $x$:

Combine like terms:

This quadratic equation in $x$ describes the $x$-coordinates of the intersection points, in terms of $m$.

A quadratic equation $ax^2 + bx + c = 0$ has:

• two distinct real solutions if $b^2 - 4ac > 0$
• exactly one real solution if $b^2 - 4ac = 0$
• no real solutions if $b^2 - 4ac < 0$

The quantity $b^2 - 4ac$ is called the discriminant.

Here, the quadratic is

So:

• $a = 1$
• $b = -(m+6)$
• $c = 12$

For exactly one solution, set the discriminant equal to $0$:

Simplify the equation

First, square the term:

Move $48$ to the other side:

Now we will solve this equation for $m$ to find the possible slopes.

From

take the square root of both sides:

Simplify $\sqrt{48}$:

So

which gives

Since $m > 0$, only $m = -6 + 4\sqrt{3}$ is allowed.

Thus, the value of $m$ is