Answer:
We can use the point-slope formula to find the equation of the tangent line. The point-slope formula is given by:
y - y1 = m(x - x1)
where (x1, y1) is a point on the line, and m is the slope of the line. To find the equation of the tangent line, we need to find the slope of the line and a point on the line.
To find the slope of the line, we can use the definition of derivative. The derivative of a function f(x) at a point x0 is defined as:
f'(x0) = lim(h->0) (f(x0 + h) - f(x0))/h
In our case, the function f(x) is given by f(x) = x^2 - 5, and the point x0 is -2, so we have:
f'(-2) = lim(h->0) (f(-2 + h) - f(-2))/h
Substituting the value of the function f(x) = x^2 - 5, we get:
f'(-2) = lim(h->0) (((-2 + h)^2 - 5) - ((-2)^2 - 5))/h
Simplifying the expression, we get:
f'(-2) = lim(h->0) (h^2 - 4h)/h
Since the limit is defined as h approaches 0, we can drop the h from the denominator since it will become insignificant in the limit:
f'(-2) = lim(h->0) (h^2 - 4h)
Now we can evaluate the limit as h approaches 0:
f'(-2) = lim(h->0) h^2 - 4h
In the limit, h^2 becomes 0, so we are left with:
f'(-2) = lim(h->0) - 4h
Since the limit is defined as h approaches 0, the value of the limit is simply -4 times 0, which is 0. Therefore, the slope of the tangent line is 0.
Now that we know the slope of the tangent line, we need to find a point on the line. We are told that the line passes through the point (-2, f(-2)) on the function f(x) = x^2 - 5. Substituting the value of the function, we get:
(-2, f(-2)) = (-2, (-2)^2 - 5) = (-2, -9)
So the point on the line is (-2, -9).
We now have the slope and a point on the line, so we can use the point-slope formula to find the equation of the line. Substituting the values into the point-slope formula, we get:
y - (-9) = 0(x - (-2))
Simplifying, we get:
y = -9
Therefore, the equation of the tangent line that passes through the point (-2, -9) with a slope of 0 is given by y = -9.