Second, plug in x = 2 to get the slope of the tangent.
Now, plug the slope and the point into the equation for the tangent line.
y − 4 = − (x − 2)That simplifies to 6x + 5y = 32. The equation of the normal line must then be
y − 4 = (x − 2)That, in turn, simplifies to −5x + 6y = 14.
PROBLEM 4. The curve y = ax^2 + bx + c passes through the point (2, 4) and is tangent to the line y = x + 1
at (0, 1). Find a, b, and c.
Answer: The curve passes through (2, 4), so if you plug in x = 2, you’ll get y = 4. Therefore,
4 = 4a + 2b + cSecond, the curve also passes through the point (0, 1), so c = 1.
Because the curve is tangent to the line y = x + 1 at (0, 1), they must both have the same slope at that
point. The slope of the line is 1. The slope of the curve is the first derivative.
= 2ax + b = 2a(0) + b = bAt (0, 1), = b. Therefore, b = 1.
Now that you know b and c, plug them back into the equation from the first step and solve for a.
4 = 4a + 2 + 1, and a = PROBLEM 5. Find the points on the curve y = 2x^3 − 3x^2 − 12x + 20 where the tangent is parallel to the x-
axis.
Answer: The x-axis is a horizontal line, so it has slope zero. Therefore, you want to know where the