The Derivatives

Question

Find the equation of the line tangent to y = -x² + 3x + 8 at x = 2

Options:

1. y = -2x + 2
2. y = -x + 12
3. y = -x + 2
4. y = -2x - 12

Answer 1

Find the equation of the line tangent to y = -x² + 3x + 8 at x = 2

A line tangent to that curve will have the same slope---and the same coordinates at the tangent point.

Find the slope using the 1st derivative:

y = -x² + 3x + 8

y' = -2x + 3

At x = 2, the slope is -1

to find the y value of the tangent point, plug in x = 2 in the original equation:

y = -(2)² + 3(2) + 8

y = 18

So the tangent point is (2,18) and the slope is -1

We need the answer in the y = mx + b format.....you have the slope (m), so just solve for b by plugging in the tangent point coordinates

Answer 2

-2x+3 is the first derivative or slope of the tangent line at a point with value x.

since x=2, then the equation of the tangent line can be found with a point slope formula.

At x=2, y=10. Just substitute 2 for x in the original quadratic equation, and solve for y.

You now have the point (2,10) and the slope m=-2(2)+3=-4+3=-1

point slope formula is y-h = m(x-k) where m=-1, h=10, and k=2

y-10=-1(x-2) which is y-10 =-x+2 or y=-x+12 which is option number 2 among your possible answers