Question

the graph of the function has one relative extreme point. plot this point and check the concavity there. using only this information, sketch the graph.
f(x)= -x² - 4x - 4
the relative extreme point on the graph is (-2,0) (type an ordered pair.)
since the value of f at the relative extreme point is , the graph is at this point

Explanation:

Step1: Find the first - derivative

Given $f(x)=-x^{2}-4x - 4$, using the power rule $(x^n)'=nx^{n - 1}$, we have $f'(x)=-2x-4$.

Step2: Find the critical points

Set $f'(x) = 0$, so $-2x-4=0$. Solving for $x$ gives $x=-2$. Substitute $x = - 2$ into $f(x)$: $f(-2)=-(-2)^{2}-4\times(-2)-4=-4 + 8-4=0$. So the relative - extreme point is $(-2,0)$.

Step3: Find the second - derivative

Differentiate $f'(x)=-2x - 4$ with respect to $x$. Using the power rule, $f''(x)=-2$.

Step4: Determine the concavity

Since $f''(-2)=-2<0$, the graph is concave down at the point $(-2,0)$.

Answer:

Since the value of $f''$ at the relative extreme point is $-2$, the graph is concave down at this point.