Find the slope and the equation of the tangent line to the graph of the function f at the specified point.
f(x)=-10/3x^2+6x+6; (-1, -10/3)
slope ??
tangent line y = ?
already answered by bobpursley
he was wrong
Kimberly, as I pointed out to you in another post, it is very important to use brackets in this format to establish the correct order of operation, since we cannot "build" fractions here.
bobpursley interpreted your equation as
f(x) = 10/(3x^2 + 6 x + 6)
and his answer is correct according to that equation
the way you typed it ....
f'(x) = (-20/3)x + 6
f'(-1) = 20/3 + 6 =38/3
so the slope is 38/3
equation:
y + 10/3 = (38/3)(x+1)
times 3
3y + 10 = 38(x+1)
3y = 38x + 38 - 10
3y = 38x + 28
y = (38/3)x + 28/3
or
38x - 3y = -28
To find the slope of the tangent line to the graph of the function f at the specified point, we need to find the derivative of the function and then substitute the x-coordinate of the given point into the derivative.
1. Find the derivative of the function f(x):
To find the derivative of f(x), we can apply the power rule and the sum/difference rule of derivatives. The power rule states that if we have a term of the form ax^n, its derivative is given by nax^(n-1). The sum/difference rule states that if we have a sum or difference of functions, the derivative of the sum/difference is the sum/difference of the derivatives.
So, applying the power rule and sum/difference rule to f(x) = (-10/3)x^2 + 6x + 6, we have:
f'(x) = (-10/3)(2x) + 6
= (-20/3)x + 6
2. Substitute the x-coordinate of the given point into f'(x) to find the slope:
We are given the point (-1, -10/3), so substitute x = -1 into f'(x):
slope = f'(-1) = (-20/3)(-1) + 6
= 20/3 + 18/3
= 38/3
Therefore, the slope of the tangent line at the point (-1, -10/3) is 38/3.
3. Find the equation of the tangent line:
The equation of a line can be expressed in the form y = mx + b, where m is the slope and b is the y-intercept.
We already found the slope to be 38/3, and we can find the corresponding y-coordinate by substituting x = -1 into the original function f(x):
y = f(-1) = (-10/3)(-1)^2 + 6(-1) + 6
= -10/3 + (-6) + 6
= -10/3 - 18/3 + 18/3
= -10/3
So, the tangent line passes through the point (-1, -10/3) and has a slope of 38/3. To find the equation of the line, we can substitute the slope and the point into the point-slope formula:
y - y1 = m(x - x1)
y - (-10/3) = (38/3)(x - (-1))
y + 10/3 = (38/3)(x + 1)
Finally, we can simplify the equation:
y + 10/3 = (38/3)x + 38/3
y = (38/3)x + 38/3 - 10/3
y = (38/3)x + 28/3
Therefore, the equation of the tangent line to the graph of the function f at the point (-1, -10/3) is y = (38/3)x + 28/3.