Question

consider the following equation of a parabola.

y ^ { 2 } - 2 y = - 8 x - 9

step 3 of 3: find three points on the parabola.

Explanation:

Step1: Complete the square for the left - hand side of the equation

We have \(y^{2}-2y=(y - 1)^{2}-1\). So the equation \(y^{2}-2y=-8x - 9\) can be rewritten as \((y - 1)^{2}-1=-8x - 9\). Then \((y - 1)^{2}=-8x-8=-8(x + 1)\).

Step2: Choose values for \(y\) and solve for \(x\)

• Case 1: Let \(y = 1\)

Substitute \(y = 1\) into \((y - 1)^{2}=-8(x + 1)\). We get \((1 - 1)^{2}=-8(x + 1)\), which simplifies to \(0=-8(x + 1)\). Solving for \(x\), we have \(x=-1\). So the point is \((-1,1)\).

• Case 2: Let \(y = 3\)

Substitute \(y = 3\) into \((y - 1)^{2}=-8(x + 1)\). Then \((3 - 1)^{2}=-8(x + 1)\), i.e., \(4=-8(x + 1)\). Divide both sides by \(-8\): \(\frac{4}{-8}=x + 1\), so \(x+1=-\frac{1}{2}\), and \(x=-\frac{3}{2}\). The point is \((-\frac{3}{2},3)\).

• Case 3: Let \(y=-1\)

Substitute \(y=-1\) into \((y - 1)^{2}=-8(x + 1)\). We have \((-1 - 1)^{2}=-8(x + 1)\), i.e., \(4=-8(x + 1)\). Divide both sides by \(-8\): \(\frac{4}{-8}=x + 1\), so \(x+1=-\frac{1}{2}\), and \(x=-\frac{3}{2}\). The point is \((-\frac{3}{2},-1)\).

Answer:

\((-1,1)\), \((-\frac{3}{2},3)\), \((-\frac{3}{2},-1)\)