Find an equation of the tangent line to the graph of the function f through the point (x0, y0) not on the graph. To find the point of tangency (x, y) on the graph of f, solve the following equation of f '(x).

f '(x) = y0 − y/x0 − x
f(x) = √x
(x0, y0) = (-16, 0)

y=?

Well, to find the point of tangency (x, y) on the graph of f, we need to solve the equation f '(x) = y0 − y/x0 − x. But before we do that, let's find the derivative of f(x) = √x.

Taking the derivative of f(x) = √x, we get:
f '(x) = (1/2) * x^(-1/2)

Now, we substitute the given values (x0, y0) = (-16, 0) into the equation f '(x) = y0 − y/x0 − x:
(1/2) * (-16)^(-1/2) = 0 - y / -16 - x

Simplifying that equation gives us:
(-1/8) = -y / (-16 - x)

To make things a bit funnier, let's multiply both sides by (-16 - x):
(-1/8) * (-16 - x) = -y

Simplifying further:
(1/8) * (16 + x) = y

So, the equation of the tangent line to the graph of f through the point (-16, 0) is:
y = (1/8) * (16 + x)

Now that we've got the answer, I hope this tangent line won't make you feel too "tangentially" confused!

To find the equation of the tangent line to the graph of the function f through the point (-16, 0), we need to find the point of tangency (x, y) on the graph of f. We are given that f(x) = √x, so the derivative of f(x) is:

f'(x) = (d/dx)√x

To find the derivative, we can use the power rule of differentiation:

f'(x) = (1/2)x^(-1/2)

Now we substitute the given values into the equation f'(x) = y0 − y/x0 − x:

(1/2)x^(-1/2) = 0 - y/(-16) - x

Multiplying both sides by -16:

-8x^(-1/2) = y + 16x

Multiplying both sides by x^(1/2):

-8 = xy^(1/2) + 16x^(3/2)

To make things easier, we can substitute y = √x:

-8 = x√x + 16x^(3/2)

Rearranging the terms:

16x^(3/2) + x√x + 8 = 0

We now have an equation that relates x and y. By solving this equation, we can find the values of x and y which satisfy the equation. However, this is a non-linear equation and finding its exact solutions may be difficult.

If you are looking for an approximate solution, you can use numerical methods such as the Newton-Raphson method or plot the equation to see where it intersects the x and y-axis.

To find the equation of the tangent line to the graph of f through the point (-16, 0), we first need to find the point of tangency (x, y) on the graph of f.

The given equation for f'(x) is f '(x) = y0 − y/x0 − x, where y0 is the y-coordinate of the point of tangency and x0 is the x-coordinate of the point of tangency. Substituting the given values, we have:

f '(x) = 0 − y/-16 − x

Now, we need to find the value of x where f'(x) is equal to 0. To do that, we set f'(x) = 0:

0 = -y/-16 - x

Next, let's solve this equation for y by multiplying both sides by -16:

0 = y + 16x

Now, isolate y:

y = -16x

So, the value of y at the point of tangency is y = -16x.

Now that we have the point of tangency (x, y) = (-16, -16x), we can find the equation of the tangent line. The equation of a line can be written in the form y = mx + b, where m is the slope of the line and b is the y-intercept.

Since the slope of the tangent line is given by the derivative f'(x), we can plug in the x-coordinate of the point of tangency (-16) into f'(x) to find the slope:

f'(-16) = y0 − y/x0 − x
f'(-16) = 0 − (-16x)/(-16) − (-16)
f'(-16) = 16x + 16

So, the slope of the tangent line at the point (-16, 0) is 16x + 16.

Now, we can use the point-slope form of a line to find the equation of the tangent line. Plug in the values of x, y, and the slope into the equation:

y - y0 = m(x - x0)
y - 0 = (16x + 16)(x - (-16))
y = (16x + 16)(x + 16)

Simplifying this equation further will give you the final equation of the tangent line.

y = x^.5

y' = .5 x^-.5

sketching this I am going to reverse slope equation

y' at tangent = m = (y-yo) /( x-xo) = .5x^-.5

y' = m = (y-0)/(x+16) = .5 x^-.5

x^.5 /(x+16) = .5 x^-.5

x/(x+16) = .5
x = .5 x + 8
.5 x = 8
x = 16
so the tangent hits the curve at x = 16
where y = 4 so at
(16,4)
so this line goes through
(-16,0) and (16,4)
You can take it from there I am sure.