Question

Which of the following is correct for the quadratic polynomial $f\left(x\right)=2x^2+8x-12$

• A. vertex $=(-2,-20)$
• B. $y-$ intercept $=(0,2)$
• C. the graph is concave downward
• D. $f\left(x\right)=0$ has no real solutions

Answer 1

The correct option is A vertex $=(-2,-20)$

Given quadratic polynomial $y=2x^2+8x-12$
where $a=2,b=8$ and $c=-12$
Since, $a>0$, so the graph of the given quadratic polynomial is concave upward.
$y-$ intercept $=(0,c)=(0,-12)$

$D=b^2-4ac=8^2-4\times 2\times (-12)=160>0$
Real and distinct roots

Vertex
$=(-\frac{b}{2a},-\frac{D}{4a})=(-\frac{8}{4},-\frac{160}{8})=(-2,-20)$