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)=7x-5x^2 at (-2,-34)
m = ??
y = ??
.
f' = 7 - 10 x
at x = -2
f'(-2) = 7 -10(-2) = 27 = m
y = 27 x + b
-34 = 27(-2) + b
b = 20
y = 27 x + 20
thank you soo much I understand it now!! :)
You are welcome :)
To find the slope of the tangent line to the graph of the function at a given point, we can use the derivative of the function. The derivative represents the slope of the tangent line at any point on the graph.
First, let's find the derivative of the function f(x)=7x-5x^2. We can use the power rule for differentiation, which states that if we have a term of the form ax^n, then its derivative is nax^(n-1).
Taking the derivative of f(x)=7x-5x^2, we get:
f'(x) = 7 - 10x
Now, substituting the x-coordinate of the given point (-2) into the derivative, we get:
f'(-2) = 7 - 10(-2)
= 7 + 20
= 27
Therefore, the slope of the tangent line at the point (-2,-34) is m = 27.
To determine the equation of the tangent line, we can use the point-slope form of a linear equation, which is given by:
y - y1 = m(x - x1)
Where (x1, y1) represents the given point and m is the slope. Substituting the values in, we get:
y - (-34) = 27(x - (-2))
y + 34 = 27(x + 2)
y + 34 = 27x + 54
y = 27x + 20
Therefore, the equation of the tangent line to the graph of the function f(x)=7x-5x^2 at the point (-2,-34) is y = 27x + 20.