what are the coordinates of the vertex of the function $f(x)=x^{2}+10x - 3$?\n(-5, -28)\n(-5, 28)\n(5…

Explanation:

Step1: Identify coefficients

For the quadratic function $f(x)=ax^{2}+bx + c$, here $a = 1$, $b=10$, $c=-3$.

Step2: Find x - coordinate of vertex

The formula for the x - coordinate of the vertex of a quadratic function is $x=-\frac{b}{2a}$. Substitute $a = 1$ and $b = 10$ into the formula: $x=-\frac{10}{2\times1}=- 5$.

Step3: Find y - coordinate of vertex

Substitute $x=-5$ into the function $f(x)=x^{2}+10x - 3$. So $f(-5)=(-5)^{2}+10\times(-5)-3=25-50 - 3=-28$.

Answer:

A. $(-5,-28)$