Find the slope and the equation of the tangent line to the graph of the function f at the specified point.
f(x) = -8/5 x^2 + 7 x + 7;(-1, -8/5)
slope
tangent line y =
To find the slope of the tangent line to the graph of a function at a specified point, we need to take the derivative of the function and then evaluate it at that point.
Given the function f(x) = -8/5 x^2 + 7x + 7, we first need to find its derivative.
To differentiate f(x), we need to apply the power rule. The power rule states that if we have a term of the form x^n, the derivative with respect to x is n*x^(n-1).
So, taking the derivative of f(x) = -8/5 x^2 + 7x + 7, we get:
f'(x) = -8/5 * 2x + 7
Simplifying, we have:
f'(x) = -16/5x + 7
To find the slope of the tangent line at the point (-1, -8/5), we substitute x = -1 into the derivative:
f'(-1) = -16/5(-1) + 7
= 16/5 + 7
= 16/5 + 35/5
= 51/5
Therefore, the slope of the tangent line at the point (-1, -8/5) is 51/5.
Now, to find the equation of the tangent line, we use the point-slope form of a linear equation:
y - y1 = m(x - x1)
where m is the slope and (x1, y1) is a point on the line. In this case, m = 51/5 and the point (-1, -8/5).
Substituting these values into the equation, we get:
y - (-8/5) = 51/5(x - (-1))
y + 8/5 = 51/5(x + 1)
y + 8/5 = 51/5x + 51/5
y = 51/5x + 51/5 - 8/5
y = 51/5x + 43/5
Hence, the equation of the tangent line to the graph of f at the point (-1, -8/5) is y = 51/5x + 43/5.