Respuesta :

well, looking at the graph, notice, its vertex is at 0,9, and it passes through the points -3,0 and 3,0, so hmm let's use 3,0, to get the equation of f(x)

[tex]\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} \boxed{y=a(x- h)^2+ k}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{}{ h},\stackrel{}{ k})\\\\ -------------------------------[/tex]

[tex]\bf vertex~(0,9)~ \begin{cases} h=0\\ k=9 \end{cases}\implies y=a(x-0)^2+9 \\\\\\ \textit{we also know that } \begin{cases} x=3\\ y=0 \end{cases}\implies 0=a(3-0)^2+9 \\\\\\ -9=9a\implies \cfrac{-9}{9}=a\implies -1=a\quad thus\quad \boxed{y=-x^2+9} \\\\\\ \left. \cfrac{dy}{dx}=-2x \right|_{x=3}\implies -6[/tex]

 From the graph attached, f'(3) = 6 will be the answer.

       From the picture attached,

Vertex of the given parabola → (0, 9)

Since, vertex form of the quadratic function is given by,

f(x) = (x - h)² + k

Here, (h, k) is the vertex of the parabola.

By substituting the vertex in the given equation,

f(x) = (x - 0)² + 9

f(x) = x² + 9

Find the derivative of the equation with respect to x,

[tex]f'(x)=\frac{d}{dx}(x^{2}+9)[/tex]

f'(x) = 2x

For x = 3,

f'(x) = 2(3)

       = 6

   Therefore, f'(x) = 6 will be the answer.

Learn more,

https://brainly.com/question/1529075