Question

The graph of an equation is given above. Identify the polynomial satisfying the graph?

• A. $y=x^2+2$
• B. $y=-x^2$
• C. $y=x^2$
• D. $y=x^2-2$

Answer 1

The correct option is C $y=x^2$

Given: The graph of an equation with four points, (-2, 4), (-1, 1), (2, 4), (1, 1)

To find the degree of the polynomial

Sol: As there are 4 points, the function becomes

$y=a{x}^3+b{x}^2+cx+d$

Replace x and y with the coordinate values given to get the system:

$(-2,4)⟹4=a(-2{)}^3+b(-2{)}^2+c(-2)+d⟹4=-8a+4b-2c+d....\left(i\right)$

$(-1,1)⟹1=a(-1{)}^3+b(-1{)}^2+c(-1)+d⟹1=-a+b-c+d.........\left(ii\right)$

$(2,4)⟹4=a(2{)}^3+b(2{)}^2+c\left(2\right)+d⟹4=8a+4b+2c+d........\left(iii\right)$

$(1,1)⟹1=a(1{)}^3+b(1{)}^2+c\left(1\right)+d⟹1=a+b+c+d...........\left(iv\right)$

$\left(i\right)+\left(iii\right)$, we get

$8b+2d=8⟹4b+d=\mathrm{4...}\left(v\right)$

$\left(ii\right)+\left(iv\right)$, we get

$2b+2d=2⟹b+d=\mathrm{1.......}\left(vi\right)$

$\left(v\right)-\left(vi\right)$, we get

$3b=3⟹b=1$

Substituting the value of $b$ in equation (vi) we get

$d=0$

Substituting the value of $b$ and $d$ in equation (iii) and (iv), we get the following new set of equation,

$8a+4\left(1\right)+2c+0=4⟹8a+2c=\mathrm{0..}\left(vii\right)$

$a+1+c+0=1⟹a+c=\mathrm{0.......}\left(viii\right)$

$\left(vii\right)-[2\times (viii\left)\right]$, we get

$6a=0⟹a=0$

Substituting the value of $a$ in equation (viii), we get

$c=0$

Hence, the equation for the given graph is

$y=0\left(x^3\right)+\left(1\right)x^2+\left(0\right)x+0⟹y=x^2$

Therefore the degree of the polynomial is 2