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write the equation of the parabola that passes through the points (-1,2…

Question

write the equation of the parabola that passes through the points (-1,2), (1,4), and (-2,7). (1 point)

$y = 4x^{2}-x - 1$
$y = 4x^{2}+x - 1$
$y = 2x^{2}-x + 1$
$y = 2x^{2}+x + 1$

Explanation:

Step1: Substitute point (-1,2) into options

For $y = 4x^{2}-x - 1$, when $x=-1$, $y=4\times(-1)^{2}-(-1)-1=4 + 1-1=4
eq2$.
For $y = 4x^{2}+x - 1$, when $x=-1$, $y=4\times(-1)^{2}+(-1)-1=4-1 - 1=2$.
For $y = 2x^{2}-x + 1$, when $x=-1$, $y=2\times(-1)^{2}-(-1)+1=2 + 1+1=4
eq2$.
For $y = 2x^{2}+x + 1$, when $x=-1$, $y=2\times(-1)^{2}+(-1)+1=2-1 + 1=2$.

Step2: Substitute point (1,4) into remaining options

For $y = 4x^{2}+x - 1$, when $x = 1$, $y=4\times1^{2}+1-1=4$.
For $y = 2x^{2}+x + 1$, when $x = 1$, $y=2\times1^{2}+1+1=2 + 1+1=4$.

Step3: Substitute point (-2,7) into remaining options

For $y = 4x^{2}+x - 1$, when $x=-2$, $y=4\times(-2)^{2}+(-2)-1=16-2 - 1=13
eq7$.
For $y = 2x^{2}+x + 1$, when $x=-2$, $y=2\times(-2)^{2}+(-2)+1=8-2 + 1=7$.

Answer:

D. $y = 2x^{2}+x + 1$