Question

in the following exercise, find the coordinates of the vertex for the parabola defined by the given quadratic function. f(x)=2x^{2}+12x + 10 the vertex is (type an ordered pair.)

Explanation:

Step1: Identify coefficients

For the quadratic function $f(x)=2x^{2}+12x + 10$, $a = 2$, $b=12$, $c = 10$.

Step2: Find x - coordinate of vertex

The formula for the x - coordinate of the vertex of a parabola $y = ax^{2}+bx + c$ is $x=-\frac{b}{2a}$. Substitute $a = 2$ and $b = 12$ into the formula: $x=-\frac{12}{2\times2}=-\frac{12}{4}=- 3$.

Step3: Find y - coordinate of vertex

Substitute $x=-3$ into the function $f(x)=2x^{2}+12x + 10$. Then $f(-3)=2\times(-3)^{2}+12\times(-3)+10=2\times9-36 + 10=18-36 + 10=-8$.

Answer:

$(-3,-8)$