Question

1. a quadratic function yields the following values: x y -4 1 -3 -5 -2 -7 -1 -5 0 1 which of the following is the equation for the function? y = - 6x² + 1 y = - 2x² - 7x + 1 y = 3x² + 9x - 11 y = 2x² + 8x + 1

Explanation:

Step1: Test the point (0, 1)

Substitute \(x = 0\) into each equation. For \(y=-6x^{2}+1\), when \(x = 0\), \(y=-6\times0^{2}+1=1\). For \(y=-2x^{2}-7x + 1\), when \(x = 0\), \(y=-2\times0^{2}-7\times0 + 1=1\). For \(y=3x^{2}+9x - 11\), when \(x = 0\), \(y=3\times0^{2}+9\times0-11=-11\). For \(y=2x^{2}+8x + 1\), when \(x = 0\), \(y=2\times0^{2}+8\times0 + 1=1\). So we can eliminate the third - option \(y = 3x^{2}+9x-11\).

Step2: Test the point (-1, -5)

Substitute \(x=-1\) into the remaining three equations. For \(y=-6x^{2}+1\), when \(x=-1\), \(y=-6\times(-1)^{2}+1=-6 + 1=-5\). For \(y=-2x^{2}-7x + 1\), when \(x=-1\), \(y=-2\times(-1)^{2}-7\times(-1)+1=-2 + 7+1=6\). For \(y=2x^{2}+8x + 1\), when \(x=-1\), \(y=2\times(-1)^{2}+8\times(-1)+1=2-8 + 1=-5\). So we can eliminate \(y=-2x^{2}-7x + 1\).

Step3: Test the point (-2, -7)

Substitute \(x=-2\) into the remaining two equations. For \(y=-6x^{2}+1\), when \(x=-2\), \(y=-6\times(-2)^{2}+1=-6\times4 + 1=-24 + 1=-23\). For \(y=2x^{2}+8x + 1\), when \(x=-2\), \(y=2\times(-2)^{2}+8\times(-2)+1=2\times4-16 + 1=8-16 + 1=-7\).

Answer:

\(y = 2x^{2}+8x + 1\)