- C First, rewrite 2x + y = k in slope-intercept form: y = −2x + k; thus, the slope of the tangent, or
the first derivative of y, is −2. Next, take the derivative of y = 2x^2 − 8x + 14 and set it equal to
−2: = 4x − 8 = −2. Use this equation to solve for the x-value that corresponds to the first
derivative of −2; the x-coordinate is . Use this x-coordinate to solve for the corresponding y-
coordinate; plug x = into the equation for the curve y. The y-coordinate equals . Finally,
go back to the equation for the tangent line and plug in the x- and y-values found. Solve for k: k
= .
- C This question is testing your knowledge of the rules of continuity, where we also discuss
differentiability.
Step 1: If the function is continuous, then plugging 1 into the top and bottom pieces of the
function should yield the same answer.
a(1 3 ) − 6(1) = b(1^2 ) + 4
a − b = b + 4
Step 2: If the function is differentiable, then plugging 1 into the derivatives of the top and
bottom pieces of the function should yield the same answer.
3 a(1^2 ) − 6 = 2b(1)
3 a − 6 = 2b
Step 3: Now we have a pair of simultaneous equations. If we solve them, we get a = −14.
- B The particle slows down when the velocity and acceleration have different signs. First, take
the first derivative of x(t) and set it equal to zero to solve for the times when the velocity is
changing sign: x′(t) = 6t^2 − 21t + 9 = 0 when t = and t = 3. Next, take the second derivative
of x(t), and determine when that is changing sign: x′′(t) = 12t − 21 = 0 when x = . Since there
are three different values for which either the velocity or acceleration equals zero, there are
four intervals to check the signs of both the velocity and acceleration:
Time Velocity Acceleration
