For the function f(x)=4x2+8 , find the equation of the tangent line to the graph of f at x=1 . If the equation of the tangent line is written as y = m x + b , what are m and b?

Answer 1

Hello Bryan,

Now, m = f'(1) = 8x = 8

Thus, y = 8x + b

f(1) = 4 + 8 =12

Tangent line passes through the point (1, 12), then 12 = 8 +b => b = 4

Answer 2

y=4x^2 + 8

slope = y' = 8x

at x = 1, slope = 8, and y= 12

the equation of the tangent line is

y=8x +b, plug in x=1 and y=12 to find b

12 =8+b

b = 4

y = 8x +4

it might help to graph the upward opening parabola and the tangent line at the point (1,12)

Answer 3

To find the slope, m, of the tangent line, you need to take the derivative of f(x) first. Be sure to use the power rule nx^n-1. If f(x) = 4x²+ 8, then f’(x) = 8x. (4(x²) becomes 4(2x) or 8x). m = 8

Plug the x-value x = 1 and y-value y = 12 into the derivative equation f’(x) = 8x, which can also be written in y = mx + b form: y = 8x + b. The values plugged in will appear as 12 = 8(1) + b. Once both sides are simplified, subtract 8 from both sides to get b = 4.

Your final tangent equation for the point (1,12) will be y = 8x + 4