Question

1. find ( y = 4x - 40 ) and ( y = -6x + 60 ) intersect. then sketch a graph of both lines and label the point of intersection.
2. remember that ( y = x^{2}-12x + 20 ) (standard form) and ( y=(x - 10)(x - 2) ) (factored form) are the same, just written in different forms.

write each equation in the other form. use a rectangle to help you.
a. ( y = x^{2}-4x - 60 )
( y=)

b. ( y=(x - 10)(2x + 5) )
( y=)

c. ( y = 4x^{2}-36 )
( y=)

d. ( y=(6x - 9)(6x - 9) )
( y=)

Explanation:

Step1: Solve for \(x\)

Since \(y = 4x-40\) and \(y=-6x + 60\), set \(4x-40=-6x + 60\).
Add \(6x\) to both sides: \(4x+6x-40=-6x+6x + 60\), which simplifies to \(10x-40 = 60\).
Add \(40\) to both sides: \(10x-40 + 40=60 + 40\), so \(10x=100\).
Divide both sides by \(10\): \(x=\frac{100}{10}=10\).

Step2: Solve for \(y\)

Substitute \(x = 10\) into \(y = 4x-40\).
\(y=4\times10-40=40 - 40=0\).

Step1: Factor \(y=x^{2}-4x - 60\)

Find two numbers \(m\) and \(n\) such that \(m + n=-4\) and \(m\times n=-60\). The numbers are \(m = 6\) and \(n=-10\).
So \(y=(x + 6)(x-10)\).

Step1: Expand \(y=(x - 10)(2x+5)\)

Use the distributive property (FOIL method):
\(y=x\times(2x)+x\times5-10\times(2x)-10\times5\)
\(y = 2x^{2}+5x-20x-50\)
\(y=2x^{2}-15x - 50\)

Step1: Factor \(y = 4x^{2}-36\)

First, factor out the GCF \(4\): \(y=4(x^{2}-9)\).
Then use the difference - of - squares formula \(a^{2}-b^{2}=(a + b)(a - b)\) with \(a=x\) and \(b = 3\).
So \(y=4(x + 3)(x-3)\)

Answer:

The point of intersection is \((10,0)\).

For problem 5:

a.