Find the slope and an equation of the tangent line to the graph of the function f at the specified point.

f(x)=-1/3x^2+5x+5: (-1, -1/3)

Answer: f’(x)=-2/3x + 5

y=-2/3x -1

I re-worked the problem and got f'(x)=5, y=5x+ 14/3

This is from a multiple choice test and this is not an answer.

Can someone check my work?

Your derivative is correct

Now use the value of x of the given point to find the slope
slope = (-2/3)(-1) + 5
= 2/3 + 5 = 17/3

equation is
y = (17/3)x + b
sub in the point
-1/3 = (17/3)(-1) + b
16/3 = b

equation: y = (17/3)x + 16/3

Thanks. I was missing the step where you plug in the -1 to get slope.

To find the slope and equation of the tangent line to the graph of the function f(x) at the point (-1, -1/3), we need to take the derivative of f(x) and substitute x = -1 to find the slope.

Given: f(x) = -1/3x^2 + 5x + 5, and the point (-1, -1/3).

Taking the derivative of f(x) using the power rule, we have:
f'(x) = (-2/3)x + 5.

Substituting x = -1 into f'(x), we get the slope of the tangent line:
f'(-1) = (-2/3)(-1) + 5 = 7/3.

So, the slope of the tangent line is 7/3.

To find the equation of the tangent line, we use the point-slope form of a linear equation:
y - y1 = m(x - x1), where (x1, y1) is the point on the line and m is the slope.

Substituting the values (-1, -1/3) and slope 7/3, we get:
y - (-1/3) = (7/3)(x - (-1)).

Simplifying, we obtain:
y + 1/3 = 7/3(x + 1).

Multiplying through by 3 to clear the fractions, we get:
3y + 1 = 7(x + 1).

Expanding, we have:
3y + 1 = 7x + 7.

Rearranging, we find the equation of the tangent line:
3y = 7x + 6.

So, the equation of the tangent line to the graph of f(x) at the point (-1, -1/3) is 3y = 7x + 6.

To find the slope and equation of the tangent line to the graph of function f at the specified point, we need to find the derivative of the function and substitute the x-coordinate of the given point into the derivative.

Given function f(x) = -1/3x^2 + 5x + 5 and the point (-1, -1/3), let's find the derivative of f(x) first.

To find the derivative, we need to apply the power rule and constant multiple rule. The power rule states that if f(x) = x^n, then f'(x) = nx^(n-1). The constant multiple rule states that if f(x) = cx, where c is a constant, then f'(x) = c.

In this case, f(x) = -1/3x^2 + 5x + 5, so we apply the power rule to each term separately. The derivative of -1/3x^2 is (-1/3) * 2x^(2-1) = -2/3x, the derivative of 5x is 5, and the derivative of 5 is 0. So, the derivative f'(x) is equal to -2/3x + 5.

Next, we substitute the x-coordinate of the given point (-1, -1/3) into the derivative f'(x) to find the slope of the tangent line at that point.

f'(-1) = -2/3(-1) + 5 = 2/3 + 5 = 17/3

So, the slope of the tangent line at the point (-1, -1/3) is 17/3.

Using the point-slope form of a line, we can write the equation of the tangent line. The point-slope form is y - y1 = m(x - x1), where (x1, y1) represents the given point and m is the slope.

Substituting the values (-1, -1/3) and the slope 17/3 into the equation, we get:

y - (-1/3) = (17/3)(x - (-1))
y + 1/3 = (17/3)(x + 1)
y + 1/3 = (17/3)x + 17/3
y = (17/3)x + 17/3 - 1/3
y = (17/3)x + 16/3

Therefore, the equation of the tangent line to the graph of the function f at the point (-1, -1/3) is y = (17/3)x + 16/3.