Question

the graph of the function has one relative extreme - point. plot this point and check the concavity there. using only the information, sketch the graph.
f(x)=-x^{2}-2x - 3
the relative extreme point on the graph is (-1,-2) (type an ordered - pair.)
since the value of f at the relative extreme point is the graph is concave down at this point

Explanation:

Step1: Find the first - derivative

Given $f(x)=-x^{2}-2x - 3$. Using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have $f^\prime(x)=-2x-2$.

Step2: Find the critical points

Set $f^\prime(x) = 0$. So, $-2x-2 = 0$. Solving for $x$ gives $x=-1$. Substitute $x = - 1$ into $f(x)$: $f(-1)=-(-1)^{2}-2(-1)-3=-1 + 2-3=-2$. The relative - extreme point is $(-1,-2)$.

Step3: Find the second - derivative

Differentiate $f^\prime(x)=-2x - 2$ with respect to $x$. Using the power rule, $f^{\prime\prime}(x)=-2$.

Step4: Determine concavity

Since $f^{\prime\prime}(x)=-2<0$ for all $x$, at the relative - extreme point $x=-1$, the value of $f^{\prime\prime}(-1)=-2$. A negative second - derivative indicates the graph is concave down.

Answer:

$-2$