the graph of the function has one relative extreme point. plot this point and check the concavity there…

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)=4x^2 - 5 the relative extreme point on the graph is (0, - 5) (type an ordered pair.) since the value of f at the relative extreme point is the graph is at this point.
Answer
Explanation:
Step1: Find the first - derivative
Differentiate $f(x)=4x^{2}-5$ using the power rule. The power rule states that if $y = ax^{n}$, then $y^\prime=anx^{n - 1}$. So, $f^\prime(x)=\frac{d}{dx}(4x^{2}-5)=8x$.
Step2: Find the critical points
Set $f^\prime(x) = 0$. So, $8x=0$, which gives $x = 0$. Substitute $x = 0$ into $f(x)$: $f(0)=4(0)^{2}-5=-5$. The relative - extreme point is $(0,-5)$.
Step3: Find the second - derivative
Differentiate $f^\prime(x)=8x$ with respect to $x$. Using the power rule, $f^{\prime\prime}(x)=\frac{d}{dx}(8x)=8$.
Step4: Determine the concavity
Evaluate $f^{\prime\prime}(x)$ at the critical point $x = 0$. Since $f^{\prime\prime}(0)=8>0$, the graph of the function $y = f(x)$ is concave up at the point $(0,-5)$.
Answer:
Since the value of $f^{\prime\prime}$ at the relative extreme point is $8$, the graph is concave up at this point.