Use the four-step process to find the slope of the tangent line to the graph of the function at the given point and determine an equation of the tangent line.f(x)= x^2 - 5x +2 (1, -2)
f'(x) = 2x-5
f'(1) = -3
So, the line with slope -3 through (1,-2) is
y+2 = -3(x-1)
Not sure what your four-step process is.
So I can just use the point-slope formula to solve equations like this?
To find the slope of the tangent line to the graph of the function at a given point, you can use the four-step process. Here's how you can apply it to find the slope and equation of the tangent line for the function f(x) = x^2 - 5x + 2 at the point (1, -2).
Step 1: Find the derivative of the function.
The derivative of f(x) = x^2 - 5x + 2 can be found using the power rule. Taking the derivative of each term separately, we get:
f'(x) = d/dx (x^2) - d/dx (5x) + d/dx (2)
= 2x - 5
Step 2: Evaluate the derivative at the given x-coordinate.
To find the slope of the tangent line at a specific point, substitute the x-coordinate of the point into the derivative. In this case, evaluate f'(x) at x = 1:
f'(1) = 2(1) - 5
= 2 - 5
= -3
So, the slope of the tangent line is -3.
Step 3: Find the y-coordinate of the given point.
The given point is (1, -2), and the y-coordinate in this point is -2.
Step 4: Use the point-slope form to write the equation of the tangent line.
The point-slope form of a linear equation is y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.
Therefore, substituting the values we have:
y - (-2) = -3(x - 1)
Simplifying, we get:
y + 2 = -3x + 3
To put the equation in slope-intercept form (y = mx + b), we subtract 2 from both sides:
y = -3x + 1
So, the equation of the tangent line to the graph of f(x) = x^2 - 5x + 2 at the point (1, -2) is y = -3x + 1.