Question

For the given quadratic equation $y=x^2-3x+2,$ select the correct option(s).

• A. Graph of $y=x^2-3x+2$ is
  
  
• B. Graph of $y=x^2-3x+2$ is
  
  
• C. $\text{vertex}\equiv (0,2);$
  $y-\text{intercept}\equiv (\frac{3}{2},-\frac{1}{4})$
• D. $\text{vertex}\equiv (\frac{3}{2},-\frac{1}{4});$
  $y-\text{intercept}\equiv (0,2)$

Answer 1

Answer: (B), (D)

$\text{vertex}\equiv (\frac{3}{2},-\frac{1}{4});$

$y-\text{intercept}\equiv (0,2)$
Given quadratic polynomial: $y=x^2-3x+2$
On comparing with the standard form of quadratic expression $\text{y}=a{x}^2+bx+c$
we get, $a=1,b=-3,c=2$
$\&D=b^2-4ac=(-3{)}^2-4\cdot 1\cdot 2=1$

Since $a>0,$ the graph would be an upward opening parabola.
Now, the vertex of the quadratic polynomial is given by:
$(-\frac{b}{2a},-\frac{D}{4a})$

$\text{Where}-\frac{b}{2a}=-\frac{-3}{2\cdot 1}=\frac{3}{2}$

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

$\therefore \text{vertex}\equiv (\frac{3}{2},-\frac{1}{4})$

Now, $y-$ intercept is given as the value of $y$ at $x=0.$
$\Rightarrow y-\text{intercept}\equiv (0,c)\equiv (0,2)$
Thus, we can draw the graph of the polynomial as: