f(t) = t2 − 1

Find the equation of the line tangent to the graph of f(t) at t = 5.
Enter the equation of the tangent line here (in terms of the variable t):
y =

I would love to help you with that, but let me warn you, I am a clown bot, not a mathematician. So, brace yourself for some clownish humor.

Now, let's find the equation of the tangent line to the graph of f(t) = t^2 - 1 at t = 5. First, we need to find the derivative of f(t) to find the slope of the tangent line.

The derivative of f(t) = t^2 - 1 is f'(t) = 2t.

Now, at t = 5, the slope of the tangent line is f'(5) = 2(5) = 10.

So, we have the slope of the tangent line, but we still need to find the y-intercept. To do that, we'll use the point-slope form of a line.

The equation of a line in point-slope form is y - y1 = m(x - x1).

Since the tangent line passes through the point (5, f(5)), we have x1 = 5 and y1 = f(5).

Plugging in the values, the equation becomes y - f(5) = 10(x - 5).

Now, all we need to do is simplify and write the equation in terms of t:

y - (5^2 - 1) = 10(t - 5).

Finally, we get y = 10t - 49.

So, the equation of the tangent line is y = 10t - 49. Enjoy your journey through the wonderful world of math with a touch of clownishness!

To find the equation of the line tangent to the graph of f(t) at t = 5, we need to find the slope of the tangent line and a point on the tangent line.

Step 1: Find the slope of the tangent line.
Since the tangent line is essentially the derivative of the function, we need to find the derivative of f(t) with respect to t.

f(t) = t^2 - 1

To find the derivative, we can use the power rule for derivatives which states that if f(t) = t^n, then f'(t) = n*t^(n-1).

Applying the power rule, we get:

f'(t) = (2)(t)^(2-1) = 2t

So, the slope of the tangent line at t = 5 is equal to the value of the derivative at t = 5, which is:

m = f'(5) = 2(5) = 10

Step 2: Find a point on the tangent line.
To find a point on the tangent line, we need to evaluate the function f(t) at t = 5.

f(5) = (5)^2 - 1 = 25 - 1 = 24

So, we have a point (5, 24) on the tangent line.

Step 3: Write the equation of the tangent line.
We now have the slope of the tangent line (m = 10) and a point on the line (5, 24). We can use the point-slope form of the equation of a line to write the equation of the tangent line.

The point-slope form is: y - y1 = m(x - x1)

Plugging in the values, we get:

y - 24 = 10(x - 5)

Now, simplify the equation:

y - 24 = 10x - 50

y = 10x - 26

Therefore, the equation of the line tangent to the graph of f(t) at t = 5 is:

y = 10x - 26

To find the equation of the line tangent to the graph of f(t) at t = 5, we need to find the derivative of f(t) and then use it to find the slope of the tangent line.

First, let's find the derivative of f(t). The derivative of t^2 is 2t, and the derivative of -1 is 0. Therefore, the derivative of f(t) is:

f'(t) = 2t

Next, we substitute t = 5 into the derivative to find the slope of the tangent line at t = 5:

f'(5) = 2(5) = 10

So, the slope of the tangent line at t = 5 is 10.

Now, we need the point (5, f(5)) on the graph of f(t) to find the equation of the tangent line. To find f(5), we substitute t = 5 into the function:

f(5) = (5^2) - 1 = 25 - 1 = 24

So, the point (5, 24) lies on the graph of f(t) at t = 5.

Now we have the slope (m = 10) and a point (5, 24) on the line. Using the point-slope form of the equation of a line, we can write the equation of the tangent line:

y - y₁ = m(x - x₁)

Substituting the values, we get:

y - 24 = 10(x - 5)

Simplifying further:

y - 24 = 10x - 50

Moving -24 to the other side:

y = 10x - 26

So, the equation of the tangent line to the graph of f(t) at t = 5 is y = 10x - 26.

I am sure you meant

f(t) = t^2 - 1
f '(t) = 2t

when t = 5
f(5) = 24
so we have the contact point (5,24)
f '(5) = 10
so the slope of the line is 10

finish it by using the method you learned to find the equation of a line, given the slope and a point