Find an equation of the tangent line to the graph at the given point. (The graph is called a serpentine.)

Question

y = ?

 

Answer 1

Hi Bella,

In order to find an equation of a line, you need a point and a slope. You are given the point (-2, -8/5).

Since the line is tangent to the curve, the slope of the line is found by taking the derivative of the function and then evaluating that derivative at the point.

(Typing the steps for the derivative is not going to look very pleasant - fair warning!)

If f(x) = (16x)/(x² + 16), then we will have to use the Quotient Rule to find the derivative.

f '(x) = [bottom (derivative of the top) - top (derivative of the bottom)] / bottom²

= [(x² + 16)(16) - (16x)(2x)] / (x² + 16)²

=[16x² + 256 - 32x²] / (x² + 16)²

= [-16x² + 256] / (x² + 16)²

Evaluating this derivative at x = -2 (which is the x-coordinate of the point you are given), this equals

f '(-2) = [-16(-2)² + 256] / ((-2)² + 16)²

= [-16(4) + 256] / (4 + 16)²

= (-64 + 256) / 20²

= 192/400 which reduces to

= 12/25

Now we are ready to find the equation of the tangent line. We can use the point-slope formula.

y - y_o = m(x - x_o)

y - (-8/5) = (12/25)(x - (-2))

y + 8/5 = (12/25)(x + 2)

y + 8/5 = (12/25)x + 24/25

y = (12/25)x + 24/25 - 8/5

y = (12/25)x + 24/25 - 40/25

y = (12/25)x - 16/25

Hope this helps!

Jim