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 = ?
f'=+10(6x+6)/(3x^2+6x+6)^2
slope at x=-1
f'=0
y=mx+b=0+b
-10/3=b
y=-10/3 is the tangent line.
that wasn't right
is f(x)=-10/(3x^2+6x+6) or is
f(x)=-(10/3x^2 )+ 6x + 6 ?
To find the slope of the tangent line to a graph at a given point, you can use the derivative of the function. The derivative gives you the rate of change of the function at any point.
First, let's find the derivative of the function f(x):
f(x) = -10/3x^2 + 6x + 6
To take the derivative, you can use the power rule and the sum rule of differentiation.
The power rule states that the derivative of x^n is n*x^(n-1).
The sum rule states that the derivative of the sum of two functions is the sum of their derivatives.
Applying the power rule and the sum rule, we differentiate each term:
f'(x) = (-10/3) * d/dx(x^2) + d/dx(6x) + 0
= (-10/3) * (2x) + 6
= -20/3 * x + 6
Now that we have the derivative, we can find the slope. The slope is equal to the value of the derivative evaluated at the x-coordinate of the given point (-1, -10/3).
Substitute x = -1 into f'(x):
slope = -20/3 * (-1) + 6
= 20/3 + 6
= 20/3 + 18/3
= 38/3
The slope of the tangent line is 38/3.
To find the equation of the tangent line, we can use the point-slope form of the equation, which states that y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.
Substituting the values into the equation, we have:
y - (-10/3) = (38/3)(x - (-1))
y + 10/3 = (38/3)(x + 1)
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 is y = (38/3)x + 28/3.