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)=6x-5x^2 at (-2, -32)
I have to use the four step process, but seem to be getting stuck...
slope=6-10x at x=-2
slope=6+20=26
that is pretty steep.
y=mx+b
-32=26*-2 + b
solve for b, and you have the tangent line equation.
y=26x+20 check the math , I did it in my head.
To find the slope m of the tangent line to the graph of the function f(x)=6x-5x^2 at the point (-2, -32) and determine an equation of the tangent line, you can follow a four-step process:
Step 1: Find the derivative of the function f(x).
The derivative of f(x) represents the slope of the tangent line at any given point on the graph. To find the derivative, you can use the power rule for differentiation. For an equation of the form f(x) = ax^n, the derivative is given by f'(x) = nax^(n-1). Applying this rule to f(x) = 6x-5x^2, the derivative is f'(x) = 6 - 10x.
Step 2: Evaluate the derivative at the given point (-2, -32).
Substitute x = -2 into the derivative f'(x) to find the slope m of the tangent line at (-2, -32). Calculate: f'(-2) = 6 - 10(-2) = 6 + 20 = 26.
Step 3: Use the slope-intercept form to find the equation of the tangent line.
The slope-intercept form of a linear equation is given by y = mx + b, where m is the slope and b is the y-intercept. We already know the slope (m = 26) from step 2. To find the y-intercept, substitute the coordinates of the given point (-2, -32) into the equation and solve for b:
-32 = 26(-2) + b
-32 = -52 + b
b = -32 + 52
b = 20
Step 4: Write the equation of the tangent line.
Now that we know the slope (m = 26) and the y-intercept (b = 20), we can write the equation of the tangent line using the slope-intercept form:
y = 26x + 20
Therefore, the equation of the tangent line to the graph of f(x) = 6x-5x^2 at the point (-2, -32) is y = 26x + 20.