Question

Y=2x^2+12x range of quadratic equation​

Answer 1

Answer: [-18, ∞)

Explanation:

Answer 2

[-18, ∞]

Explanation:

Y=2x^2+12x

Case 1

a>0 then f(x)≥c−b*b/4a.

In fact for any y≥c−b/2a we have

So the range of f(x) is [c−(b* b) /4a, ∞)

Case 2

a<0 then f(x)≤c−b*b/4a and for any y<c−b/2a

So the range of f(x) is [−∞, c−(b* b) /4a]

So, for this, a=2>0, b=12 and c=0;

Applying case 1, because a=2

Range=[ c−(b* b) /4a, ∞]

=[0-(12*12)/(4*2), ∞]

=[0-144/8, ∞]

=[0-18, ∞]

=[-18, ∞]