f(t) = − 4 t2 − t− 6

Find the equation of the line tangent to the graph of f(t) at t = 8.
Enter the equation of the tangent line here (in terms of the variable t):

y = -4 t^2 -t - 6

slope = dy/dt = -8 t - 1
at t = 8, slope = -65

so our tangent line is of form
y = -65 t + b

to find b, we need a point on the line
when t = 8,
y = -4(64) -8 -6
= -270

so use that point (8, -270) to find b
-270 = -65(8) + b
b = 250
so in the end
y = -65 t + 250

Well, finding the equation of a tangent line can be quite challenging, but luckily for you, I'm here to give it a funny shot!

Let's first find the derivative of f(t) with respect to t, which we'll call f'(t). Using the power rule, we have:

f'(t) = -8t - 1

Now that we have the derivative, we can find the slope of the tangent line at t = 8 by plugging t = 8 into f'(t):

f'(8) = -8(8) - 1
= -65

So, the slope of the tangent line is -65.

Next, we need a point on the graph of f(t) to find the y-intercept of the tangent line. Let's find f(8):

f(8) = -4(8)^2 - 8 - 6
= -262

Therefore, the tangent line passes through the point (8, -262).

Using the point-slope form of a linear equation, we can finally write the equation of the tangent line as:

y - (-262) = -65(t - 8)

Simplifying this equation, we get:

y + 262 = -65t + 520
y = -65t + 258

So, the equation of the tangent line at t = 8 is y = -65t + 258.

And there you have it! An equation as straight as a clown's wig!

To find the equation of the line tangent to the graph of f(t) at t = 8, we need to find the derivative of f(t) and evaluate it at t = 8.

Step 1: Find the derivative of f(t).
To find the derivative of f(t), we need to differentiate each term individually.

The derivative of −4t^2 is (-4)*2t = -8t.

The derivative of -t is -1.

The derivative of -6 is 0 (since it is a constant).

Therefore, the derivative of f(t) is f'(t) = -8t - 1.

Step 2: Evaluate the derivative at t = 8.
To find the slope of the tangent line at t = 8, we evaluate the derivative at t = 8.

f'(8) = -8(8) - 1 = -65.

Step 3: 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 and b is the y-intercept.

We already know the slope of the tangent line is -65.

To find the y-intercept, we need a point on the tangent line. Since the tangent line passes through the function f(t) at t = 8, we can find the corresponding y-value by substituting t = 8 into f(t).

f(8) = -4(8)^2 - 8 - 6 = -278.

So, the point (8, -278) lies on the tangent line.

Using the point-slope form, we can write the equation of the tangent line:

y - (-278) = -65(t - 8).

Simplifying this equation will give us the final equation of the tangent line.

To find the equation of the tangent line to the graph of f(t) at t = 8, we need to find the derivative of f(t) and evaluate it at t = 8.

First, let's find the derivative of f(t) = -4t^2 - t - 6.

To find the derivative, we can apply the power rule and the constant rule.

The power rule states that if we have a function of the form f(t) = t^n, then the derivative is given by f'(t) = n*t^(n-1).

Applying the power rule to each term in f(t):

f'(t) = -4*(2t^(2-1)) - 1*(t^(1-1)) - 0

Simplifying, we have:

f'(t) = -8t - 1

Now, we can evaluate f'(t) at t = 8:

f'(8) = -8(8) - 1 = -65

Now, we have the slope of the tangent line at t = 8.

The equation of a line in point-slope form is given by y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.

Since we are interested in finding the equation in terms of the variable t, we will use the form t - t1 = m(t - t1).

We have the point (8, f(8)) on the graph of f(t), so (t1, f(t1)) = (8, f(8)).

Substituting the values into the equation:

t - 8 = -65(t - 8)

Expanding and simplifying:

t - 8 = -65t + 520

Combining like terms:

66t = 528

Dividing by 66:

t = 8

Therefore, the equation of the tangent line in terms of the variable t is:

t - 8 = -65(t - 8)