Question

Which quadratic equation fits the data in the table?

$x$	$y$
$-5$	$33$
$-2$	$9$
$-1$	$5$
$0$	$3$
$3$	$9$
$4$	$15$
$6$	$33$

• A. $\mathrm{y}=-{\mathrm{x}}^2-\mathrm{x}-3$
• B. $\mathrm{y}={\mathrm{x}}^2-\mathrm{x}+3$
• C. $\mathrm{y}={\mathrm{x}}^2-\mathrm{x}-3$
• D. $\mathrm{y}={\mathrm{x}}^2+\mathrm{x}+3$

Answer 1

The correct option is B

$\mathrm{y}={\mathrm{x}}^2-\mathrm{x}+3$

Explanation for correct option:

Option(B): $\mathrm{y}={\mathrm{x}}^2-\mathrm{x}+3$

Given that for $\mathrm{x}=0$, $\mathrm{y}=3$

Substituting in the given equation, we get,

$3=0-0+3$

$\Rightarrow$ $3=3$

$\Rightarrow$L.H.S.$=$R.H.S.

Now, for $\mathrm{x}=-1$, $\mathrm{y}=5$

Substituting in the given equation, we get,

$5={\left(-1\right)}^2-\left(-1\right)+3$

$\Rightarrow$ $5=1+1+3$

$\Rightarrow$ $5=5$

$\Rightarrow$L.H.S.$=$R.H.S.

Explanation for incorrect options:

Option(A): $\mathrm{y}=-{\mathrm{x}}^2-\mathrm{x}-3$

Given that for $\mathrm{x}=0$, $\mathrm{y}=3$

Substituting in the given equation, we get,

$3=-0-0-3$

$\Rightarrow$ $3\ne -3$

$\Rightarrow$L.H.S.$\ne$R.H.S.

Option(C): $\mathrm{y}={\mathrm{x}}^2-\mathrm{x}-3$

Given that for $\mathrm{x}=0$, $\mathrm{y}=3$

Substituting in the given equation, we get,

$3=0-0-3$

$\Rightarrow$ $3\ne -3$

$\Rightarrow$L.H.S.$\ne$R.H.S.

Option(D): $\mathrm{y}={\mathrm{x}}^2+\mathrm{x}+3$

Given that for $\mathrm{x}=0$, $\mathrm{y}=3$

Substituting in the given equation, we get,

$3=0+0+3$

$\Rightarrow$ $3=3$

$\Rightarrow$L.H.S.$=$R.H.S.

Now, for $\mathrm{x}=-1$, $\mathrm{y}=5$

Substituting in the given equation, we get,

$5={\left(-1\right)}^2+\left(-1\right)+3$

$\Rightarrow$ $5=1-1+3$

$\Rightarrow$ $5\ne 3$

$\Rightarrow$L.H.S.$\ne$R.H.S.

Hence, the correct option is option(B).