for the function $f(x)=x^{2}-5$, find the equation of the tangent line at $x = 6$.
for the function $f(x)=x^{2}-5$, find the equation of the tangent line at $x = 6$.
Answer
Explanation:
Step1: Find the derivative of the function
The function is (f(x) = x^{2}-5). Using the power rule ((x^{n})^\prime=nx^{n - 1}), the derivative (f^\prime(x)=2x).
Step2: Find the slope of the tangent line at (x = 6)
Substitute (x = 6) into the derivative. The slope (m=f^\prime(6)=2\times6 = 12).
Step3: Find the (y)-coordinate of the point on the function at (x = 6)
Substitute (x = 6) into (f(x)). (y=f(6)=6^{2}-5=36 - 5=31). So the point ((x_0,y_0)=(6,31)).
Step4: Use the point - slope form of a line (y - y_0=m(x - x_0))
Substitute (m = 12), (x_0 = 6) and (y_0 = 31) into the formula. (y-31=12(x - 6)). Expand the right - hand side: (y-31=12x-72). Add 31 to both sides to get the slope - intercept form: (y=12x- 41).
Answer:
The equation of the tangent line is (y = 12x-41)