tangent line in y = mx + b format

Question

hi! can someone help be find out the tangent line for f(x) = x^6 at x = 3 in a y = mx + b format?

Answer 1

First get f '(x) using the power rule.

f '(x) = 6x⁵

Therefore to get the slope of the tangent line at x = 3, get f '(3).

f '(3) = 6 * (3)⁵ = 1458

It would be best to use point slope form of a line to start.

At x = 3, f(x) = (3)⁶ = 729. Therefore, the tangent goes through the point (3, 729).

We have a point and a slope of the tangent, so but it into point-slope form.

y - y₁ = m(x - x₁)

y - 729 = 1458(x - 3)

Now just but this into a y = mx + b format.

Answer 2

Hello, Lily,

The slope of a line can be determined by taking it's first derivative and solving for x=3. The first derivative of a function roduces an equation that describes the rate of increase of that function at a point given by x.

f(x) = x⁶

f(x)' = 6x⁵

Slope at x = 3 is 1458 (6*3⁵= 1458)

This line has the format y=mx+b, where m is the slope. So we have y = 1458x + b, but we still need a value for b. Do that by finding the one point we know is on this line,by solving f(x) for x=3, and then using that point to solve for b.

f(x) = x⁶

f(3) = (3)⁶

f(3) = 729

Use this point (3, 729) in the equation y = 1458x+b and solve for b:

729=1458*(3)+b

b = -3645

The line tangent to y=x⁶ is y=1458x-3645

I hope I'm correct,

Bob