Question

in the following exercise, find the coordinates of the vertex for the parabola defined by the given quadratic function. f(x)=3x^2 + 6x + 10 the vertex is (type an ordered pair.)

Explanation:

Step1: Identify coefficients

For the quadratic function $f(x)=3x^{2}+6x + 10$, we have $a = 3$, $b=6$, $c = 10$.

Step2: Find x - coordinate of vertex

Use the formula $x=-\frac{b}{2a}$. Substitute $a = 3$ and $b = 6$ into it: $x=-\frac{6}{2\times3}=- 1$.

Step3: Find y - coordinate of vertex

Substitute $x=-1$ into the function $f(x)=3x^{2}+6x + 10$. So $f(-1)=3\times(-1)^{2}+6\times(-1)+10=3 - 6 + 10=7$.

Answer:

$(-1,7)$