Question

Select the correct statements for the quadratic polynomial $y=x^2+5x+6.$

• A. $y_{min}=-\frac{5}{2}$
• B. Graph of $y=x^2+5x+6$ is:
  
• C. $y-\text{intercept}\equiv (0,6)$
• D. $\text{Vertex}\equiv (-\frac{5}{2},-\frac{1}{4})$

Answer 1

Answer: (C), (D)

$\text{Vertex}\equiv (-\frac{5}{2},-\frac{1}{4})$

Given: $y=x^2+5x+6$
On comparing with standard form of quadratic expression $\text{y}=a{x}^2+bx+c$
We get: $a=1,b=5,c=6$
$\&D=b^2-4ac=(5{)}^2-4\cdot 1\cdot 6=1$

Now, the vertex of the quadratic polynomial is given by:
$V\equiv (-\frac{b}{2a},-\frac{D}{4a})$

$\text{Where}-\frac{b}{2a}=-\frac{5}{2.1}=-\frac{5}{2}$

$\&-\frac{D}{4a}=-\frac{1}{4.1}=-\frac{1}{4}$

$\therefore \text{Vertex}\equiv (-\frac{5}{2},-\frac{1}{4})$

Now, $y-$ intercept is given as the value of $y$ for $x=0$
$\Rightarrow y-\text{intercept}\equiv (0,6)$
We know, for an upward opening parabola, we only have minimum value and that minimum value is at vertex.
$\therefore y_{min}=-\frac{1}{4}$