Consider the parabola y = 8x − x2. (a) find the slope of the tangent line to the parabola at the point (1, 7). 6 (b) find an equation of the tangent line in part (a)

The slope of the tangent line to a point in any given function is equal to the derivate of the function evaluated in that point.

The solutions are:

Our function is y = f(x) = 8*x - x^2

a) Here we need to derivate our function, we will get:

f'(x) = 8 - 2*x

To get the slope of the tangent line at the point (1, 7) we need to evaluate the derivated function at the same x-value of the point, we will get:

x = 1

Then:

f'(1) = 8 - 2*1 = 8 - 2 = 6

The slope is 6.

b) We want to find the equation of the line.

Remember that a general line can be written as:

y = a*x + b

where a is the slope and b is the y-intercept.

Here we already know the slope, so we have a = 6.

Then our line becomes:

y = 6*x + b

To find the value of b we can use the fact that this line passes through the point (1, 7)

This means that when x = 1, we have y = 7.

Replacing these in the line equation we get:

7 = 6*1 + b

7 = 6 + b

7 - 6 = b

1 = b

Then the equation of the line is:

y = 6*x + 1

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a) 6

b)

Step-by-step explanation:

Suppose we have a function y = f(x).

The slope of the tangent line at the point x = x0 is

In this problem, we have that:

(a) find the slope of the tangent line to the parabola at the point (1, 7).

So

(b) find an equation of the tangent line in part (a)

We have a point

The equation for the tangent line is:

So