Question

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)

Answer 1

In order to find any of the info we need we have to find the first derivative of the equation.  If $y=-x^2+8x$,  then  $y'=-2x+8$.  We are told to find the slope at point (1, 7).  Using that x value in our derivative will give us the slope of the line at that point.  y' = -(1)^2+8.  So y' = 7.  That's the slope of the line.  Now we will use that slope along with the x and y coordinate in the slope-intercept form of a line to solve for b.  7 = 7(1) + b so b = 0.  Our equation then is y = 7x.  If you graph these in the same window on your calculator, you can see how perfectly that align at the given point.  It's really quite perfect.

Answer 2

Answer:

a) 6

b) $y = 6x + 1$

Step-by-step explanation:

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

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

$m = y'(x0)$

In this problem, we have that:

$y = 8x - x^{2}$

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

$x_{0} = 1$

So

$m = y'(1) = 8 - 2x = 8 - 2 = 6$

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

We have a point $(x_{0}, y_{0})$

The equation for the tangent line is:

$y - y_{0} = m(x - x_{0})$

So

$x_{0} = 1, y_{0} = 7$

$y - y_{0} = m(x - x_{0})$

$y - 7 = 6(x - 1)$

$y = 6x - 6 + 7$

$y = 6x + 1$