Find the slope m of the tangent line to the graph of the function at the given point and determine an equation of the tangent line.
f(x) = 7 x - 2 x^2 text( at ) \(-1,-9\)
m =
y =
dy/dx = 7 - 4x
at (-1,-9)
dy/dx = m = 7 - 4(-1) = 11
plug in point
-9 = 11 (-1) + b
-9 = -11 + b
b = 2
For the slope, calculate df/dx at x = -1.
m = df/dx = 7 - 14x = 21
y = 21x + b
-9 = -21 + b
b = 12
y = 21x +12
To find the slope of the tangent line to the function at the given point \((-1, -9)\), we first need to find the derivative of the function \(f(x) = 7x - 2x^2\).
Differentiating the function, we get:
\[\frac{dy}{dx} = 7 - 4x\]
Substitute the value \(x = -1\) into the derivative to find the slope:
\[\frac{dy}{dx} = 7 - 4(-1) = 7 + 4 = 11\]
So, the slope of the tangent line is \(m = 11\).
To find the equation of the tangent line, we can use the point-slope form of the line equation:
\[y - y_1 = m(x - x_1)\]
Using the given point \((-1, -9)\) and the slope \(m = 11\),
\[y - (-9) = 11(x - (-1))\]
Simplifying,
\[y + 9 = 11(x + 1)\]
Expanding,
\[y + 9 = 11x + 11\]
Subtracting 9 from both sides,
\[y = 11x + 2\]
Therefore, the equation of the tangent line is \(y = 11x + 2\).
To find the slope (m) of the tangent line to the graph of the function at the given point, we can use the first derivative of the function evaluated at that point.
First, let's find the first derivative of the function f(x):
f'(x) = 7 - 4x
Next, let's evaluate the derivative at the point (-1, -9):
f'(-1) = 7 - 4(-1)
= 7 + 4
= 11
So, the slope (m) of the tangent line is 11.
To determine an 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 given point.
Using (-1, -9), we have:
y - (-9) = 11(x - (-1))
y + 9 = 11(x + 1)
y + 9 = 11x + 11
y = 11x + 2
Therefore, the equation of the tangent line is y = 11x + 2.