Answer:
[tex] m_1 = m_2 = \frac{1}{4}[/tex]
And then since the new equation have the following form:
[tex] y = m_2 x +b [/tex]
We can use the point given (x=12, y = -25) in order to find the intercept with this equation:
[tex] -25 = \frac{1}{4} (12) +b[/tex]
And solving for the intercept b we got:
[tex] -25 = 3 +b[/tex]
We subtract in both sides 3 and we got:
[tex] b = -25-3 = -28[/tex]
And our final equation who satisfy the condition given is:
[tex] y= \frac{1}{4} x -28[/tex]
Step-by-step explanation:
For this case we have the following equation given:
[tex] y = \frac{1}{4} x + 2[/tex]
And we want to find an equation of a line parallel to the given function and this case we need to satisfy this condition:
[tex] m_1 = m_2 = \frac{1}{4}[/tex]
And then since the new equation have the following form:
[tex] y = m_2 x +b [/tex]
We can use the point given (x=12, y = -25) in order to find the intercept with this equation:
[tex] -25 = \frac{1}{4} (12) +b[/tex]
And solving for the intercept b we got:
[tex] -25 = 3 +b[/tex]
We subtract in both sides 3 and we got:
[tex] b = -25-3 = -28[/tex]
And our final equation who satisfy the condition given is:
[tex] y= \frac{1}{4} x -28[/tex]
