Question

What is the range of the function f(x)=3x^2+12x + 18

Answer 1

Answer: {y∈R: y≤6} or [6,∞)

Explanation:

This problem doesn't require too much math. If you look at the equation given, you can see that it is a quadratic equation in the form of$y=Ax^2+Bx+C$. Since this is a quadratic equation, we have an idea of that the graph would look like. It either curves up or down. Since this is a positive equation, $(+3x^2)$, we know that this is going to curve up. In order to find the minimum of the curve, you would use $\frac{-b}{2a}$.

$\frac{-12}{2(3)}=-2$

This means the x value of the parabola is -2. To find the y, you plug -2 into the  original equation.

$f(2)=3(-2)^2+12(-2)+18$

$f(2)=6$

Now that we know the y value of the minimum/vertex is 6, and it is determined that the parabola curves up, the range is y≤6 because the range starts at 6 and goes off toward infinity.