at what point do the graph of the function f(x)=(x^2)+4x-1 have a horizontal tangent?
That is a parabola. It is horizontal at the vertex.
Of course you could do it the sophisticated way, where the derivative is zero
f' = 2 x + 4
= 0 when x = -2
then
y = -5
To find the point on the graph of the function f(x) = x^2 + 4x - 1 where it has a horizontal tangent, we need to find the values of x where the slope of the tangent line is equal to 0. In other words, we need to find the x-values where the derivative of the function is equal to 0.
Step 1: Find the derivative of the function f(x).
The derivative of f(x) is denoted as f'(x) or dy/dx, where y = f(x).
In this case, f(x) = x^2 + 4x - 1.
To find the derivative, we can use the power rule and add up the derivatives of each term:
f'(x) = 2x + 4.
Step 2: Set the derivative equal to 0 and solve for x.
To find the x-values where the slope is 0, we set f'(x) = 0:
2x + 4 = 0.
Subtract 4 from both sides:
2x = -4.
Divide both sides by 2:
x = -2.
Step 3: Find the corresponding y-value.
To find the corresponding y-value on the graph, substitute the x-value into the original function:
f(-2) = (-2)^2 + 4(-2) - 1
= 4 - 8 - 1
= -5.
Therefore, the graph of the function f(x) = x^2 + 4x - 1 has a horizontal tangent at the point (-2, -5).