The line $y = 2x + c$ is tangent to the parabola $y^2 = 8x.$  Find $c.$
Rearranging $y = 2x + c$ gives $2x = y - c.$  Substituting into $y^2 = 8x,$ we get
\[y^2 = 4(y - c) = 4y - 4c,\]or $y^2 - 4y + 4c = 0.$  Since we have a tangent, this quadratic will have a double root.  In other words, its discriminant will be 0.  Hence, $(-4)^2 - 4(4c) = 16 - 16c = 0,$ which means $c = \boxed{1}.$
1
Level 2
Intermediate Algebra
Problem: The line $y = 2x + c$ is tangent to the parabola $y^2 = 8x.$  Find $c.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the value of \( c \) such that the line \( y = 2x + c \) is tangent to the parabola \( y^2 = 8x \), we need to set up the condition for tangency. 

First, we substitute the equation of the line into the equation of the parabola. This gives us:

\[
(2x + c)^2 = 8x
\]

Expanding the left side, we have:

\[
4x^2 + 4cx + c^2 = 8x
\]

Rearranging this equation leads to:

\[
4x^2 + (4c - 8)x + c^2 = 0
\]

For the line to be tangent to the parabola, this quadratic equation in \( x \) must have exactly one solution. This occurs when the discriminant is zero. The discriminant \( D \) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by \( D = b^2 - 4ac \). Here, \( a = 4 \), \( b = 4c - 8 \), and \( c = c^2 \).

Calculating the discriminant:

\[
D = (4c - 8)^2 - 4 \cdot 4 \cdot c^2
\]

Simplifying this expression:

\[
D = (16c^2 - 64c + 64) - 16c^2
\]
\[
D = -64c + 64
\]

Setting the discriminant equal to zero for tangency:

\[
-64c + 64 = 0
\]

Solving for \( c \):

\[
-64c = -64
\]
\[
c = 1
\]

Thus, the value of \( c \) that makes the line tangent to the parabola is \( 1 \).<|END_OF_LATENT|> The final answer is $1$.