Given that the graph of f passes through the point (1, -4) and that the slope of its tangent line at (x,f(x)) is 2x+1, find f(-1) (graph is hypothetical).
well, dy/dx = 2x+1
so y = x^2 + x + c
but (1,-4) lies on it, so
-4 = 1 + 1 + c
c = -6
f(x) = x^2 + x - 6
find f(-1) = ....
I had gotten -6 as an answer earlier but 2nd guessed it, thanks.
To find f(-1), we need to integrate the given slope function and apply the initial condition.
Step 1: Integrate the given slope function:
To integrate 2x+1, we obtain:
∫(2x+1) dx = x^2 + x + C.
Step 2: Apply the initial condition:
Given that f passes through the point (1, -4), we substitute the values into the integrated function:
-4 = (1^2 + 1 + C).
Solving for C:
-4 = (1 + 1 + C),
-4 = 2 + C,
C = -6.
Therefore, the integrated function is f(x) = x^2 + x - 6.
Step 3: Evaluate f(-1):
Substituting x = -1 into the function:
f(-1) = (-1)^2 + (-1) - 6,
f(-1) = 1 - 1 - 6,
f(-1) = -6.
Hence, f(-1) = -6.
To find f(-1) using the given information, we can use the concept of the slope of the tangent line.
Step 1: Start by finding the equation of the tangent line at (1, -4) using the given slope.
The slope of the tangent line is given as 2x+1. Since we want the tangent line to pass through the point (1, -4), we can substitute the values of x and y into the slope equation to find the value of the y-intercept (b).
Given: slope (m) = 2x + 1
Point: (1, -4)
Using the point-slope form of a linear equation, we have:
(y - y1) = m(x - x1)
Substituting the values:
(y - (-4)) = (2(1) + 1)(x - 1)
(y + 4) = (2 + 1)(x - 1)
(y + 4) = 3(x - 1)
Expanding and simplifying:
y + 4 = 3x - 3
y = 3x - 7
Therefore, the equation of the tangent line is y = 3x - 7.
Step 2: To find f(-1), we need to find the value of y when x = -1 on the graph of f.
Substitute x = -1 into the equation y = 3x - 7:
y = 3(-1) - 7
y = -3 - 7
y = -10
Therefore, f(-1) = -10.