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)