Question

consider this! find the points of intersection of the two given graphs using calculation. (\begin{cases}y = x^{2}-axcdots\textcircled{1}\\y = bx - abcdots\textcircled{2}end{cases}) sol substituting (\textcircled{1}) into (\textcircled{2}), (x^{2}-ax=bx - ab), (x^{2}-ax - bx+ab = 0), (x^{2}-(a + b)x+ab = 0), ((x - a)(x - b)=0), (x=a,b). when (x = a), from (\textcircled{1}), (y = 0). when (x = b), from (\textcircled{1}), (y=b^{2}-ab). so, ((a,0),(b,b^{2}-ab))

Explanation:

Step1: Set the two equations equal

Since at the intersection - points, the \(y\) - values of the two functions are equal, we set \(x^{2}-2x + 2=x^{2}-8x + 20\).

Step2: Simplify the equation

Subtract \(x^{2}\) from both sides of the equation:
\(x^{2}-2x + 2-x^{2}=x^{2}-8x + 20-x^{2}\), which simplifies to \(-2x + 2=-8x+20\).

Step3: Solve for \(x\)

Add \(8x\) to both sides: \(-2x+8x + 2=-8x+8x + 20\), getting \(6x + 2 = 20\).
Then subtract 2 from both sides: \(6x+2 - 2=20 - 2\), so \(6x=18\).
Divide both sides by 6: \(x = 3\).

Step4: Solve for \(y\)

Substitute \(x = 3\) into the first equation \(y=x^{2}-2x + 2\).
\(y=3^{2}-2\times3 + 2=9-6 + 2=5\).

Answer:

The point of intersection is \((3,5)\)