Find the slope of the tangent to the graph of the function y = 8x ^ -3 / 4 + 3 at the point x0 = 1.

We have a function:

y = 8 * x ^ (- 3/4) + 3.

Let us find the slope of the tangent drawn to the graph of the function at the point with the abscissa x0 = 1.

We find the slope of the tangent from the formula itself of the tangent to the graph of the function. Let’s write this equation:

y = y ‘(x0) * (x – x0) + y (x0).

The slope of the tangent is the coefficient of the variable in the equation of the straight line (k in the formula y = k * x + b).

As you can see, in the tangent equation this coefficient is y ‘(x0). Let’s find this value:

y ‘(x) = 8 * (-3/4) * x ^ (- 7/4) = -6 * x ^ (- 7/4).

y ‘(x0) = -6 * 1 ^ (- 7/4) = -6.