Let f(x)=(x^2)+x+14. What is the value of f(x) for which the tangent line to the graph of y=f(x) is parallel to the x-axis?
In other words, where is the derivative zero or where is the vertex.
f' = 2x + 1
0 = 2 x + 1
x = -1/2
check by completing the square
x^2 + x = - 14
x^2 + x + 1/4 = -14 + 1/4
(x+1/2)^2 = - 13 3/4
sure enough, vertex at x = -1/2
f(-1/2) = 1/4 -1/2 + 14
= 13 1/4
remarkable :)
To find the value of x for which the tangent line to the graph of y = f(x) is parallel to the x-axis, we need to find the derivative of the function f(x). The tangent line to a graph is parallel to the x-axis when the derivative of the function is equal to zero.
Let's start by finding the derivative of f(x):
f(x) = x^2 + x + 14
To find the derivative, we differentiate each term of the function:
f'(x) = (2x) + (1) + 0
Simplifying the derivative, we have:
f'(x) = 2x + 1
Now, we set the derivative equal to zero:
2x + 1 = 0
Subtracting 1 from both sides:
2x = -1
Dividing both sides by 2:
x = -1/2
Therefore, the value of x for which the tangent line to the graph of y = f(x) is parallel to the x-axis is x = -1/2.