Find the slope of the tangent line to the graph of the function at the given point.
g(x) = 17 − x^2; (3, 8)
g'(x) = -2x
at(3,8) , slope = -6
y-8 = -6(x-3)
re-arrange to any form you are familiar with.
Thnks :)
To find the slope of the tangent line to the graph of the function g(x) = 17 - x^2 at the point (3, 8), we need to find the derivative of the function and evaluate it at x = 3.
Step 1: Find the derivative of the function g(x) = 17 - x^2.
The derivative of a function is denoted by g'(x) or dy/dx. For g(x) = 17 - x^2, we can apply the power rule of differentiation, which states that if f(x) = x^n, then f'(x) = nx^(n-1).
So, using the power rule, we have:
g'(x) = d/dx (17 - x^2)
= 0 - 2x
= -2x
Step 2: Evaluate the derivative at x = 3.
Plugging x = 3 into the derivative, we get:
g'(3) = -2(3)
= -6
Therefore, the slope of the tangent line to the graph of g(x) = 17 - x^2 at the point (3, 8) is -6.
To find the slope of the tangent line to the graph of the function at the given point (3, 8), we need to find the derivative of the function and evaluate it at x = 3.
Step 1: Find the derivative of g(x).
The given function is g(x) = 17 - x^2. To find the derivative, we differentiate each term separately. The derivative of a constant is zero, so the derivative of 17 is zero. The derivative of x^2 is 2x. Therefore, the derivative of g(x) = 17 - x^2 is g'(x) = 0 - 2x, or simply g'(x) = -2x.
Step 2: Evaluate the derivative at x = 3.
Now that we have the derivative, g'(x) = -2x, we can substitute x = 3 into the derivative to find the slope at that point.
Therefore, g'(3) = -2(3) = -6.
Hence, the slope of the tangent line to the graph of the function g(x) = 17 - x^2 at the point (3, 8) is -6.