Find the vertex of the parabola
-$f(x)=3 x^{2}-60 x+301$
A) $(10,0)$
B) $(1,0)$
C) $(10,1)$
D) $(1,10)$

Question

Quizplus · Accepted Answer

The vertex of a parabola in the form $f(x) = ax^2 + bx + c$ is given by the formula $(h, k)$, where $h = -\frac{b}{2a}$. For $f(x) = 3x^2 - 60x + 301$, $a = 3$ and $b = -60$. Thus, $h = -\frac{-60}{2 \times 3} = 10$. To find $k$, substitute $x = 10$ into the function: $f(10) = 3(10)^2 - 60(10) + 301 = 1$. Therefore, the vertex is $(10, 1)$.

Find the Vertex of the Parabola - f(x)=3x2−60x+301f(x)=3 x^{2}-60 x+301f(x)=3x2−60x+301 A) (10,0)(10,0)(10,0)

Find the Vertex of the Parabola
- $f(x)=3 x^{2}-60 x+301$ A) $(10,0)$